∫ WBCHSE - Three Dimentional Coordinate Geometry
50+ Three Dimentional Coordinate Geometry Marks · 2M 4M
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📘 2 Mark Integrals
30+ problemsSolution:
We find the lengths of the sides.
\(AB = \sqrt{(2-0)^{2}+(-1-1)^{2}+(3-2)^{2}}\)
\(= \sqrt{4+4+1} = \sqrt{9} = 3\)
\(BC = \sqrt{(1-2)^{2}+(-3+1)^{2}+(1-3)^{2}}\)
\(= \sqrt{1+4+4} = \sqrt{9} = 3\)
\(AC = \sqrt{(1-0)^{2}+(-3-1)^{2}+(1-2)^{2}}\)
\(= \sqrt{1+16+1} = \sqrt{18} = 3\sqrt{2}\)
Since \(AB = BC\), the triangle is isosceles.
Now,
\(AB^{2} + BC^{2} = 9 + 9 = 18\)
\(= AC^{2}\)
Hence, the triangle satisfies Pythagoras theorem.
\(\therefore\) The triangle is an isosceles right angled triangle.
Solution:
Let the yz-plane divide the line joining \(A(4,8,10)\) and \(B(6,10,-8)\) in the ratio \(m:n\).
The x–coordinate of the dividing point is \(\frac{m(6) + n(4)}{m+n}\)
Since it is the yz-plane, so \(x=0\).
Therefore, \( \frac{6m + 4n}{m+n} = 0\)
\(\Rightarrow 6m + 4n = 0\)
\(\Rightarrow 3m + 2n = 0\)
\(\Rightarrow \frac{m}{n} = -\frac{2}{3}\)
Hence, the ratio is \(2:3\).
\(\therefore\) The yz-plane divides the line internally in the ratio \(2:3\).
Solution:
The equation of the yz-plane is \(x=0\).
Reflection in the yz-plane changes the sign of the x-coordinate only.
The y and z coordinates remain unchanged.
Given point is \((-3, 4, 3)\).
So, on changing the sign of x-coordinate, we get
\((3, 4, 3)\)
\(\therefore\) The required image point is \((3, 4, 3)\).
Solution:
The equation of the xy-plane is \(z=0\).
Perpendicular projection on the xy-plane makes the z-coordinate zero.
The x and y coordinates remain unchanged.
Given point is \((3,1,2)\).
So, on putting \(z=0\), we get
\((3,1,0)\)
\(\therefore\) The required projection is \((3,1,0)\).
Solution:
Let the required point on the Z-axis be \((0,0,z)\).
Distance from \((0,0,z)\) to \((1,5,7)\) is
\(\sqrt{(0-1)^2 + (0-5)^2 + (z-7)^2}\)
\(= \sqrt{1 + 25 + (z-7)^2}\)
Distance from \((0,0,z)\) to \((5,1,-4)\) is
\(\sqrt{(0-5)^2 + (0-1)^2 + (z+4)^2}\)
\(= \sqrt{25 + 1 + (z+4)^2}\)
Since the point is equidistant,
\(\therefore 1 + 25 + (z-7)^2 = 25 + 1 + (z+4)^2\)
\(\Rightarrow (z-7)^2 = (z+4)^2\)
\(\Rightarrow z^2 - 14z + 49 = z^2 + 8z + 16\)
\(\Rightarrow -22z = -33\)
\(\Rightarrow z = \frac{33}{22} = \frac{3}{2}\)
\(\therefore\) The required point is \(\left(0,0,\frac{3}{2}\right)\).
Solution:
For point \((-3,1,2)\):
\(x<0,\; y>0,\; z>0\)
This corresponds to the 2nd octant.
For point \((-3,3,-3)\):
\(x<0,\; y>0,\; z<0\)
This corresponds to the 6th octant.
\(\therefore\) The first point lies in the 2nd octant and the second point lies in the 6th octant.
Solution:
Let the required point be \((a,b,c)\).
Distance from \((a,b,c)\) to \((0,0,0)\) is
\(\sqrt{a^{2}+b^{2}+c^{2}}\)
Distance from \((a,b,c)\) to \((x,0,0)\) is
\(\sqrt{(a-x)^{2}+b^{2}+c^{2}}\)
Since the point is equidistant,
\(a^{2}+b^{2}+c^{2}=(a-x)^{2}+b^{2}+c^{2}\)
\(\Rightarrow a^{2}=(a-x)^{2}\)
\(\Rightarrow a=\frac{x}{2}\)
Similarly, equating distance from \((0,0,0)\) and \((0,y,0)\),
\(b=\frac{y}{2}\)
Equating distance from \((0,0,0)\) and \((0,0,z)\),
\(c=\frac{z}{2}\)
\(\therefore\) The required point is \(\left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)\).
Solution:
Let one vertex of the unit cube be at the origin \((0,0,0)\).
Since the sides coincide with the rectangular axes and each side is 1 unit, the opposite vertex is \((1,1,1)\).
The length of the diagonal is the distance between \((0,0,0)\) and \((1,1,1)\).
\(= \sqrt{(1-0)^{2}+(1-0)^{2}+(1-0)^{2}}\)
\(= \sqrt{1+1+1}\)
\(= \sqrt{3}\)
\(\therefore\) The length of the diagonal of the unit cube is \(\sqrt{3}\) units.
Solution:
Let the vertices of the parallelogram be \(A(x_1,y_1,z_1)\), \(B(x_2,y_2,z_2)\), \(C(x_3,y_3,z_3)\) and \(D(x_4,y_4,z_4)\).
In a parallelogram, diagonals bisect each other.
Mid-point of diagonal \(AC\) is
\(\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}, \frac{z_1+z_3}{2}\right)\)
Mid-point of diagonal \(BD\) is
\(\left(\frac{x_2+x_4}{2}, \frac{y_2+y_4}{2}, \frac{z_2+z_4}{2}\right)\)
Since the diagonals bisect each other, their mid-points are equal.
\(\Rightarrow \frac{x_1+x_3}{2}=\frac{x_2+x_4}{2}\)
\(\Rightarrow x_1+x_3=x_2+x_4\)
Similarly,
\(y_1+y_3=y_2+y_4\)
\(z_1+z_3=z_2+z_4\)
Hence Proved.
Solution:
Since the point P lie on X-axis, point Q lie on Y-axis and point R lie on Z-axis,
Therefore \(P(p,0,0), \quad Q(0,q,0), \quad R(0,0,r)\)
Let the required point be \((a,b,c)\).
Distance from \((a,b,c)\) to origin \(O(0,0,0)\) is
\(\sqrt{a^{2}+b^{2}+c^{2}}\)
Distance from \((a,b,c)\) to \(P(p,0,0)\) is
\(\sqrt{(a-p)^{2}+b^{2}+c^{2}}\)
Since the point is equidistant from O and P,
\(a^{2}+b^{2}+c^{2}=(a-p)^{2}+b^{2}+c^{2}\)
\(\Rightarrow a^{2}=(a-p)^{2}\)
\(\Rightarrow a=\frac{p}{2}\)
Similarly, equating distance from O and Q,
\(b=\frac{q}{2}\)
Equating distance from O and R,
\(c=\frac{r}{2}\)
\(\therefore\) The required point is \(\left(\frac{p}{2},\frac{q}{2},\frac{r}{2}\right)\).
Solution:
Coordinates of \(P\) are \((3,12,4)\).
Hence, \(\overrightarrow{OP} = 3\hat{i} + 12\hat{j} + 4\hat{k}\).
Magnitude of \(\overrightarrow{OP}\) is
\(|\overrightarrow{OP}| = \sqrt{3^{2}+12^{2}+4^{2}}\)
\(= \sqrt{9+144+16}\)
\(= \sqrt{169}\)
\(= 13\)
Direction cosines are
\(l=\frac{3}{13}, \quad m=\frac{12}{13}, \quad n=\frac{4}{13}\)
\(\therefore\) The direction cosines of \(\overrightarrow{OP}\) are \(\left(\frac{3}{13}, \frac{12}{13}, \frac{4}{13}\right)\).
Solution:
Given direction ratios of \(\overrightarrow{OP}\) are \((1,-2,-2)\).
Hence, coordinates of \(P\) are proportional to \((1,-2,-2)\).
Let \(P = (k, -2k, -2k)\).
Since distance from origin is 3 units,
\(\sqrt{k^{2} + (-2k)^{2} + (-2k)^{2}} = 3\)
\(\Rightarrow \sqrt{k^{2} + 4k^{2} + 4k^{2}} = 3\)
\(\Rightarrow \sqrt{9k^{2}} = 3\)
\(\Rightarrow 3|k| = 3\)
\(\Rightarrow k = \pm 1\)
Therefore,
If \(k=1\), \(P = (1,-2,-2)\)
If \(k=-1\), \(P = (-1,2,2)\)
\(\therefore\) The coordinates of \(P\) are \((1,-2,-2)\) or \((-1,2,2)\).
Solution:
Perpendicular projections on the coordinate axes are proportional to the direction ratios of the line.
Hence, direction ratios (d.r.'s) are \( (12, 4, 13) \).
Magnitude of the line is
\(\sqrt{12^{2} + 4^{2} + 13^{2}}\)
\(= \sqrt{144 + 16 + 169}\)
\(= \sqrt{329}\)
Therefore, direction cosines are
\(l = \frac{12}{\sqrt{329}}, \quad m = \frac{4}{\sqrt{329}}, \quad n = \frac{13}{\sqrt{329}}\)
\(\therefore\) The direction cosines are \(\left(\frac{12}{\sqrt{329}}, \frac{4}{\sqrt{329}}, \frac{13}{\sqrt{329}}\right)\).
Solution:
Let \(O(0,0,0)\) and \(P(2,3,-1)\).
Direction ratios of the line \(OP\) are
\((2-0,\, 3-0,\, -1-0)\)
\(= (2, 3, -1)\)
Magnitude of \(OP\) is
\(\sqrt{2^{2} + 3^{2} + (-1)^{2}}\)
\(= \sqrt{4 + 9 + 1}\)
\(= \sqrt{14}\)
Therefore, direction cosines are
\(l = \frac{2}{\sqrt{14}}, \quad m = \frac{3}{\sqrt{14}}, \quad n = \frac{-1}{\sqrt{14}}\)
\(\therefore\) The direction cosines of the required line are \(\left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)\).
Solution:
For direction cosines \(l, m, n\),
\(l^{2} + m^{2} + n^{2} = 1\)
Given \(l = \frac{1}{2}, \quad m = \frac{1}{3}\)
\(\Rightarrow \left(\frac{1}{2}\right)^{2} + \left(\frac{1}{3}\right)^{2} + n^{2} = 1\)
\(\Rightarrow \frac{1}{4} + \frac{1}{9} + n^{2} = 1\)
\(\Rightarrow \frac{9 + 4}{36} + n^{2} = 1\)
\(\Rightarrow \frac{13}{36} + n^{2} = 1\)
\(\Rightarrow n^{2} = 1 - \frac{13}{36}\)
\(\Rightarrow n^{2} = \frac{36 - 13}{36}\)
\(\Rightarrow n^{2} = \frac{23}{36}\)
\(\Rightarrow n = \pm \frac{\sqrt{23}}{6}\)
\(\therefore\) The value of \(n\) is \(\pm \frac{\sqrt{23}}{6}\).
Solution:
Given symmetric form of the line is
\(\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}\)
Let each ratio be equal to \(t\).
\(\Rightarrow \frac{x-5}{3} = t\)
\(\Rightarrow x = 5 + 3t\)
\(\Rightarrow \frac{y+4}{7} = t\)
\(\Rightarrow y = -4 + 7t\)
\(\Rightarrow \frac{6-z}{2} = t\)
\(\Rightarrow 6 - z = 2t\)
\(\Rightarrow z = 6 - 2t\)
Hence, coordinates of a point on the line are \( (5,-4,6) \)
Direction ratios are \( (3,7,-2) \)
\(\Rightarrow\) Vector equation of the line is
\(\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + t(3\hat{i} + 7\hat{j} - 2\hat{k})\)
\(\therefore\) Required vector equation is \(\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + t(3\hat{i} + 7\hat{j} - 2\hat{k})\).
Solution:
Given vector equation of the line is
\(\vec{r} = (2\hat{i} + \hat{j} - 4\hat{k}) + \lambda(\hat{i} - \hat{j} - \hat{k})\)
Comparing with \( \vec{r} = \vec{a} + \lambda \vec{b} \),
Point on the line \(= (2,1,-4)\)
Direction ratios \(= (1,-1,-1)\)
Therefore, cartesian (symmetric) form is
\(\frac{x-2}{1} = \frac{y-1}{-1} = \frac{z+4}{-1}\)
\(\therefore\) Required cartesian equation is \(\frac{x-2}{1} = \frac{y-1}{-1} = \frac{z+4}{-1}\).
Solution:
Given \(x = y = z\).
Let \(x = y = z = t\).
\(\Rightarrow x = t, \quad y = t, \quad z = t\)
Hence, coordinates of a point on the line are \( (0,0,0) \)
Direction ratios are \( (1,1,1) \)
Hence, vector form is
\(\vec{r} = t(\hat{i} + \hat{j} + \hat{k})\)
\(\therefore\) Required vector equation is \(\vec{r} = t(\hat{i} + \hat{j} + \hat{k})\).
Solution:
Given equations are
\(x = 1\), \(\frac{y-2}{3} = \frac{z+1}{2}\)
Let \(\frac{y-2}{3} = \frac{z+1}{2} = t\).
\(\Rightarrow y-2 = 3t \Rightarrow y = 2 + 3t\)
\(\Rightarrow z+1 = 2t \Rightarrow z = -1 + 2t\)
Also, \(x = 1\) i.e. \(x = 1 + 0 t\) (constant)
\(\Rightarrow\) Direction ratios are \( (0, 3, 2) \).
\(\therefore\) Required direction ratios are \((0, 3, 2)\).
Solution:
For all values of \(m\) and \(n\), the lines are perpendicular.
\(\Rightarrow\) Their direction vectors are perpendicular.
Direction vector of first line:
\(\vec{b}_1 = (5, -\lambda, -3)\)
Direction vector of second line:
\(\vec{b}_2 = (-\lambda, -\lambda, 2)\)
Since both the given lines are perpendicular,
So, \(\vec{b}_1 \cdot \vec{b}_2 = 0\)
\(\Rightarrow (5)(-\lambda) + (-\lambda)(-\lambda) + (-3)(2) = 0\)
\(\Rightarrow -5\lambda + \lambda^{2} - 6 = 0\)
\(\Rightarrow \lambda^{2} - 5\lambda - 6 = 0\)
\(\Rightarrow (\lambda - 6)(\lambda + 1) = 0\)
\(\Rightarrow \lambda = 6 \text{ or } -1\)
\(\therefore\) The value of \( \lambda \) is \(6\) or \(-1\).
Solution:
The line passes through origin \(O(0,0,0)\).
Since it is parallel to the x-axis,
\(\Rightarrow\) Direction ratios are \( (1,0,0) \).
Vector equation of the line is
\(\vec{r} = t( \hat{i} ) + 0( \hat{j} ) + 0( \hat{k} )\)
Cartesian (parametric) form is
\(x = t,\quad y = 0,\quad z = 0\)
\(\therefore\) Required equation of the line is \(x = t,\; y = 0,\; z = 0\).
Solution:
For the first line:
\(x = ay + b,\quad z = cy + d\)
Let \(y = t\)
\(\Rightarrow x = at + b,\quad y = t,\quad z = ct + d\)
\(\Rightarrow\) Direction ratios are \( (a, 1, c) \)
For the second line:
\(x = a'y + b',\quad z = c'y + d'\)
Let \(y = s\)
\(\Rightarrow x = a's + b',\quad y = s,\quad z = c's + d'\)
\(\Rightarrow\) Direction ratios are \( (a', 1, c') \)
Since lines are perpendicular,
\(\Rightarrow (a,1,c)\cdot(a',1,c') = 0\)
\(\therefore aa' + 1 + cc' = 0\) (Hence Proved)
Solution:
Direction ratios of first line:
\((3, -16, 7)\)
Direction ratios of second line:
\((3, 8, -5)\)
The required line is perpendicular to both lines.
\(\Rightarrow\) Its direction ratios are perpendicular to both.
\(\Rightarrow\) Required direction ratios are cross product of the two given direction vectors.
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \)
\(= \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))\)
\(= \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)\)
\(= 24\hat{i} + 36\hat{j} + 72\hat{k}\)
\(\Rightarrow\) Direction ratios \(= (24, 36, 72)\)
\(\Rightarrow\) Simplified d.r.'s \(= (2, 3, 6)\)
Since line passes through \( (1,2,-4) \),
Vector equation is
\(\vec{r} = ( \hat{i} + 2\hat{j} - 4\hat{k} ) + t(2\hat{i} + 3\hat{j} + 6\hat{k})\)
\(\therefore\) Required equation is \(\vec{r} = ( \hat{i} + 2\hat{j} - 4\hat{k} ) + t(2\hat{i} + 3\hat{j} + 6\hat{k})\).
Solution:
Given line is
\(\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}\)
\(\Rightarrow\) Direction ratios are \( (3, 5, 6) \)
The required line is parallel to the given line.
\(\Rightarrow\) It has the same direction ratios \( (3, 5, 6) \)
Since it passes through \( (1,2,3) \),
Vector equation is
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(3\hat{i} + 5\hat{j} + 6\hat{k})\)
\(\therefore\) Required equation is \(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(3\hat{i} + 5\hat{j} + 6\hat{k})\).
Solution:
Let direction cosines of the line be
\(l = \cos \alpha,\quad m = \cos \beta,\quad n = \cos \gamma\)
We know that for direction cosines,
\(l^{2} + m^{2} + n^{2} = 1\)
Using Cauchy–Schwarz inequality,
\((1+1+1)(l^{2}+m^{2}+n^{2}) \ge (l+m+n)^{2}\)
\(\Rightarrow 3 \cdot 1 \ge (l+m+n)^{2}\)
\(\Rightarrow 3 \ge (l+m+n)^{2}\)
\(\Rightarrow l+m+n \le \sqrt{3}\)
\(\Rightarrow \cos \alpha + \cos \beta + \cos \gamma \le \sqrt{3}\)
\(\therefore\) Proved.
Solution:
From first line:
\(\frac{2x-2}{2} = \frac{3y+3}{3\lambda} = \frac{z+5}{2}\)
\(\Rightarrow \frac{x-1}{1} = \frac{y+1}{\lambda} = \frac{z+5}{2}\)
\(\Rightarrow\) Direction ratios are \( (1, \lambda, 2) \)
From second line:
\(\frac{x-2}{3} = \frac{y+1}{-2\lambda} = \frac{2z-1}{0}\)
\(\Rightarrow\) Direction ratios are \( (3, -2\lambda, 0) \)
Since lines are perpendicular,
\(\Rightarrow (1,\lambda,2) \cdot (3,-2\lambda,0) = 0\)
\(\Rightarrow 3 - 2\lambda^{2} + 0 = 0\)
\(\Rightarrow 3 - 2\lambda^{2} = 0\)
\(\Rightarrow 2\lambda^{2} = 3\)
\(\Rightarrow \lambda^{2} = \frac{3}{2}\)
\(\Rightarrow \lambda = \pm \sqrt{\frac{3}{2}}\)
\(\therefore\) Required value of \( \lambda \) is \(\pm \sqrt{\frac{3}{2}}\).
Solution:
Given direction ratios:
\(\vec{a} = (1, 2, -3)\)
\(\vec{b} = (0, 1, 2)\)
The required line is perpendicular to both.
\(\Rightarrow\) Its direction ratios are \( \vec{a} \times \vec{b} \).
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 0 & 1 & 2 \end{vmatrix} \)
\(= \hat{i}(2 \cdot 2 - (-3)\cdot 1) - \hat{j}(1 \cdot 2 - (-3)\cdot 0) + \hat{k}(1 \cdot 1 - 2 \cdot 0)\)
\(= \hat{i}(4 + 3) - \hat{j}(2 - 0) + \hat{k}(1 - 0)\)
\(= 7\hat{i} - 2\hat{j} + \hat{k}\)
\(\Rightarrow\) Direction ratios are \( (7, -2, 1) \)
Magnitude \(= \sqrt{7^{2} + (-2)^{2} + 1^{2}}\)
\(= \sqrt{49 + 4 + 1} = \sqrt{54}\)
\(\Rightarrow\) Direction cosines are
\(\left( \frac{7}{\sqrt{54}}, \frac{-2}{\sqrt{54}}, \frac{1}{\sqrt{54}} \right)\)
\(\therefore\) Required direction cosines are \(\left( \frac{7}{\sqrt{54}}, \frac{-2}{\sqrt{54}}, \frac{1}{\sqrt{54}} \right)\).
Solution:
Let \(Q\) divide \(PR\) in the ratio \(m:n\).
Using section formula,
\( Q = \left( \frac{m x_R + n x_P}{m+n}, \frac{m y_R + n y_P}{m+n}, \frac{m z_R + n z_P}{m+n} \right) \)
Given \(P(1,2,3)\), \(R(7,8,9)\), \(Q(4,5,6)\)
\(\Rightarrow 4 = \frac{7m + 1n}{m+n}\)
\(\Rightarrow 4m + 4n = 7m + n\)
\(\Rightarrow 3n = 3m\)
\(\Rightarrow m = n\)
Similarly from y and z coordinates we get the same result.
\(\Rightarrow m:n = 1:1\)
\(\therefore\) \(Q\) divides \(PR\) internally in the ratio \(1:1\).
Solution:
If the line makes equal angles with the axes,
\(\Rightarrow l = m = n\)
Let \(l = m = n = k\).
We know,
\(l^{2} + m^{2} + n^{2} = 1\)
\(\Rightarrow k^{2} + k^{2} + k^{2} = 1\)
\(\Rightarrow 3k^{2} = 1\)
\(\Rightarrow k^{2} = \frac{1}{3}\)
\(\Rightarrow k = \pm \frac{1}{\sqrt{3}}\)
\(\therefore\) Required direction cosines are \(\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)\) (or all negative).
📙 4 Mark Integrals
28+ problemsSolution:
As PS is the angle bisector of \( \angle QPR \), so we have
\(\Rightarrow \dfrac{QS}{SR} = \dfrac{PQ}{PR}\)
\(PQ = \sqrt{(5-2)^2 + (6-2)^2 + (9+3)^2}\)
\(= \sqrt{3^2 + 4^2 + 12^2}\)
\(= \sqrt{9 + 16 + 144} = \sqrt{169} = 13\)
\(PR = \sqrt{(2-2)^2 + (7-2)^2 + (9+3)^2}\)
\(= \sqrt{0 + 5^2 + 12^2}\)
\(= \sqrt{25 + 144} = \sqrt{169} = 13\)
\(\therefore \dfrac{QS}{SR} = \dfrac{13}{13} = 1:1\)
So, \( QS = SR \)
\(\therefore S\) is midpoint of \(QR\).
Coordinates of \(S\):
\( S = \left( \frac{5+2}{2}, \frac{6+7}{2}, \frac{9+9}{2} \right) \)
\(= \left(\frac{7}{2}, \frac{13}{2}, 9 \right)\)
Now the length of \(PS\)is -
\(PS = \sqrt{\left(\frac{7}{2}-2\right)^2 + \left(\frac{13}{2}-2\right)^2 + (9+3)^2}\)
\(= \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{9}{2}\right)^2 + 12^2}\)
\(= \sqrt{\frac{9}{4} + \frac{81}{4} + 144}\)
\(= \sqrt{\frac{90}{4} + 144}\)
\(= \sqrt{\frac{90 + 576}{4}}\)
\(= \sqrt{\frac{666}{4}}\)
\(= \frac{\sqrt{666}}{2}\)
\(\therefore\) Coordinates of \(S\) are \(\left(\frac{7}{2}, \frac{13}{2}, 9 \right)\) and \( |PS| = \) \(\frac{\sqrt{666}}{2}\).
Solution:
Given relations:
\(l + m + n = 0\)
\(l^2 + m^2 - n^2 = 0\)
Also, since \(l, m, n\) are direction cosines,
\(\therefore l^2 + m^2 + n^2 = 1\)
From \(l^2 + m^2 - n^2 = 0\),
\(\Rightarrow l^2 + m^2 = n^2\)
Substitute in \(l^2 + m^2 + n^2 = 1\), we get
\(\Rightarrow n^2 + n^2 = 1\)
\(\Rightarrow 2n^2 = 1\)
\(\Rightarrow n^2 = \frac{1}{2}\)
Now, we have \( l^2 + m^2 = n^2\), So
\(\Rightarrow l^2 + m^2 = \frac{1}{2}\)
Now on squaring \(l + m + n = 0\), we get
\(l^2 + m^2 + n^2 + 2(lm + mn + nl) = 0\)
\(\Rightarrow 1 + 2(lm + mn + nl) = 0\)
\(\Rightarrow lm + mn + nl = -\frac{1}{2}\)
Let the required angle be \(\theta\).
Therefore, we can write
\(\cos \theta = lm + mn + nl\)
\(\Rightarrow \cos \theta = -\frac{1}{2}\)
Since acute angle is required,
\(\theta = 60^\circ\)
\(\therefore\) The required acute angle is \(60^\circ\).
(a) the lines are parallel if \( \dfrac{a^2}{u} + \dfrac{b^2}{v} + \dfrac{c^2}{w} = 0 \),
(b) the lines are perpendicular if \( a^2(v+w) + b^2(w+u) + c^2(u+v) = 0 \).
We have, \( al + bm + cn = 0 \)
\(\Rightarrow l = -\dfrac{bm + cn}{a}\)
Now, \( ul^2 + vm^2 + wn^2 = 0 \).
\(\Rightarrow u\left(\dfrac{bm + cn}{a}\right)^2 + vm^2 + wn^2 = 0\)
\(\Rightarrow u(bm + cn)^2 + va^2m^2 + wa^2n^2 = 0\)
\(\Rightarrow u(b^2m^2 + c^2n^2 + 2bcmn) + va^2m^2 + wa^2n^2 = 0\)
\(\Rightarrow (ub^2 + va^2)m^2 + 2ubcmn + (uc^2 + wa^2)n^2 = 0\)
\(\Rightarrow (ub^2 + va^2)\left(\dfrac{m}{n}\right)^2 + 2ubc\left(\dfrac{m}{n}\right) + (uc^2 + wa^2) = 0 \quad ...(i)\)
Let its roots are \( \dfrac{m_1}{n_1} \) and \( \dfrac{m_2}{n_2} \).
(a) If two straight lines are parallel, the Eq. (i) will provide us equal roots.
Thus, \( D = 0 \)
\(\Rightarrow (2ubc)^2 - 4(ub^2 + va^2)(uc^2 + wa^2) = 0\)
\(\Rightarrow (ubc)^2 - (b^2u + a^2v)(c^2u + a^2w) = 0\)
\(\Rightarrow (b^2u + a^2v)(c^2u + a^2w) - (ubc)^2 = 0\)
\(\Rightarrow a^2b^2uw + a^2c^2uv + a^4vw = 0\)
\(\Rightarrow b^2uw + c^2uv + a^2vw = 0\)
\(\Rightarrow \dfrac{a^2}{u} + \dfrac{b^2}{v} + \dfrac{c^2}{w} = 0\)
(b) When two straight lines are perpendicular, the product of the roots \(= \dfrac{uc^2 + wa^2}{ub^2 + va^2}\)
\(\Rightarrow \dfrac{m_1}{n_1} \cdot \dfrac{m_2}{n_2} = \dfrac{uc^2 + wa^2}{ub^2 + va^2}\)
\(\Rightarrow \dfrac{m_1 m_2}{uc^2 + wa^2} = \dfrac{n_1 n_2}{ub^2 + va^2}\)
Similarly, eliminating \(n\), we get
\(\Rightarrow \dfrac{l_1 l_2}{wb^2 + vc^2} = \dfrac{m_1 m_2}{uc^2 + wa^2}\)
Thus,
\(\dfrac{l_1 l_2}{wb^2 + vc^2} = \dfrac{m_1 m_2}{uc^2 + wa^2} = \dfrac{n_1 n_2}{ub^2 + va^2}\)
Two lines are perpendicular, if
\(l_1l_2 + m_1m_2 + n_1n_2 = 0\)
\(\Rightarrow (wb^2 + vc^2) + (uc^2 + wa^2) + (ub^2 + va^2) = 0\)
\(\Rightarrow a^2(u+v) + b^2(u+w) + c^2(v+w) = 0\)
Hence, the result.
Solution:
Let the side of the cube be \(a\).
Take one vertex of the cube as origin \(O(0,0,0)\).
Then three adjacent vertices are
\(A(a,0,0),\; B(0,a,0),\; C(0,0,a)\).
One body diagonal is
\(\vec{d_1} = \overrightarrow{OC} = (a,a,a)\)
Another body diagonal is
\(\vec{d_2} = (a,-a,a)\)
(joining opposite vertices)
Now, we have
\(\vec{d_1}\cdot\vec{d_2}\)
\(= a(a) + a(-a) + a(a)\)
\(= a^2 - a^2 + a^2\)
\(= a^2\)
Magnitude of each diagonal is the following:
\(|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}\)
\(|\vec{d_2}| = a\sqrt{3}\)
Let angle between two body diagonals be \( \theta \).
\( \cos\theta = \dfrac{\vec{d_1}\cdot\vec{d_2}}{|\vec{d_1}||\vec{d_2}|} \)
\( = \dfrac{a^2}{(a\sqrt{3})(a\sqrt{3})} \)
\( = \dfrac{a^2}{3a^2} \)
\( = \dfrac{1}{3} \)
\(\therefore\) \( \theta = \cos^{-1}\!\left(\dfrac{1}{3}\right) \).
Solution:
Let direction cosines of the line be
\(l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma\)
We know that,
\(l^2 + m^2 + n^2 = 1\)
\(\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \quad ...(1)\)
(i)
\(\sin^2\alpha = 1 - \cos^2\alpha\)
\(\sin^2\beta = 1 - \cos^2\beta\)
\(\sin^2\gamma = 1 - \cos^2\gamma\)
On adding, we get
\( \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) \)
\( = 3 - 1 \)
\( = 2 \)
\(\therefore\) \( \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2 \,\text{(Hence Proved)}\).
(ii)
\(\cos 2\alpha = 2\cos^2\alpha - 1\)
\(\cos 2\beta = 2\cos^2\beta - 1\)
\(\cos 2\gamma = 2\cos^2\gamma - 1\)
On adding, we get
\( \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) - 3 \)
\( = 2(1) - 3 \)
\( = -1 \)
Now we have, \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma + \sin^2\alpha + \sin^2\beta + \sin^2\gamma = -1 + 2 \)
\(\therefore\) \( \cos 2\alpha + \cos 2\beta + \cos 2\gamma + \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 1 \, \text{(Hence Proved)}\).
Solution:
Given,
\(l - 5m + 3n = 0 \quad ...(1)\)
\(7l^2 + 5m^2 - 3n^2 = 0 \quad ...(2)\)
From (1),
\(l = 5m - 3n\)
Substitute in (2):
\(7(5m - 3n)^2 + 5m^2 - 3n^2 = 0\)
\(\Rightarrow 7(25m^2 + 9n^2 - 30mn) + 5m^2 - 3n^2 = 0\)
\(\Rightarrow 175m^2 + 63n^2 - 210mn + 5m^2 - 3n^2 = 0\)
\(\Rightarrow 180m^2 + 60n^2 - 210mn = 0\)
\(\Rightarrow 6m^2 + 2n^2 - 7mn = 0\)
\(\Rightarrow 6m^2 - 3mn - 4mn + 2n^2 = 0\)
\(\Rightarrow 3m(2m - n) - 2n(2m - n) = 0\)
\(\Rightarrow (3m - 2n)(2m - n) = 0\)
\(\Rightarrow 3m - 2n = 0 \quad \text{or} \quad 2m - n = 0\)
Case 1: \(3m - 2n = 0\)
\(\Rightarrow 3m = 2n\)
\(\Rightarrow \dfrac{m}{n} = \dfrac{2}{3}\)
Let \(m = 2k,\; n = 3k\)
\(l = 5m - 3n = 10k - 9k = k\)
\(\Rightarrow\) Direction ratios \((l,m,n) = (1,2,3)\)
Magnitude \(= \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\)
Hence, Direction cosines are \( \left( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right) \)
Case 2: \(2m - n = 0\)
\(\Rightarrow 2m = n\)
\(\Rightarrow \dfrac{m}{n} = \dfrac{1}{2}\)
Let \(m = k,\; n = 2k\)
\(l = 5m - 3n = 5k - 6k = -k\)
\(\Rightarrow\) Direction ratios \((l,m,n) = (-1,1,2)\)
Magnitude \(= \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{6}\)
Hence, Direction cosines are \( \left( -\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \)
\(\therefore\) Required direction cosines are
\( \left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right) \) and \( \left(-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right) \)
Solution:
Given,
\(l + m + n = 0 \quad ...(1)\)
\(lm = 0 \quad ...(2)\)
Since \(lm = 0\), either \(l = 0\) or \(m = 0\).
Case 1: \(l = 0\)
From (1),
\(m + n = 0 \Rightarrow m = -n\)
Using \(l^2 + m^2 + n^2 = 1\),
\(0 + m^2 + n^2 = 1\)
\(m^2 + (-m)^2 = 1\)
\(2m^2 = 1\)
\(m^2 = \frac{1}{2}\)
\(m = \pm \frac{1}{\sqrt{2}}, \quad n = \mp \frac{1}{\sqrt{2}}\)
Thus direction cosines of first line are
\( \left(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) \)
or
\( \left(0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \)
Case 2: \(m = 0\)
From (1),
\(l + n = 0 \Rightarrow l = -n\)
Using \(l^2 + m^2 + n^2 = 1\),
\(l^2 + 0 + n^2 = 1\)
\(l^2 + (-l)^2 = 1\)
\(2l^2 = 1\)
\(l = \pm \frac{1}{\sqrt{2}}, \quad n = \mp \frac{1}{\sqrt{2}}\)
Thus direction cosines of second line are
\( \left(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}\right) \)
or
\( \left(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right) \)
Now angle between the two lines is given by -
\( \cos\theta = l_1l_2 + m_1m_2 + n_1n_2 \)
\( = 0\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\cdot 0 + \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) \)
\( = \frac{1}{2} \)
\(\Rightarrow \theta = \cos^{-1}\left(\frac{1}{2}\right)\)
\(\Rightarrow \theta = 60^\circ\)
\(\therefore\) The required angle is \(60^\circ\).
Solution:
Given line:
\(\dfrac{x-2}{-3} = \dfrac{y+3}{1} = \dfrac{z+3}{6}\)
Direction ratios of the given line are
\((-3,\, 1,\, 6)\)
The required line is parallel to the given line.
\(\Rightarrow\) It has same direction ratios \( (-3,1,6) \).
Since it passes through \( (1,2,3) \),
Cartesian (symmetric) form of the required line is
\(\dfrac{x-1}{-3} = \dfrac{y-2}{1} = \dfrac{z-3}{6}\)
Vector form:
Position vector of point \( (1,2,3) \) is \((\hat{i} + 2\hat{j} + 3\hat{k})\)
Hence vector equation is
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-3\hat{i} + \hat{j} + 6\hat{k})\)
\(\therefore\) Required equation is
\(\dfrac{x-1}{-3} = \dfrac{y-2}{1} = \dfrac{z-3}{6}\)
and vector form
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-3\hat{i} + \hat{j} + 6\hat{k})\) .
Solution:
Given line:
\(\dfrac{-x-2}{1} = \dfrac{y+3}{7} = \dfrac{2z-6}{3}\)
Rewrite in standard symmetric form.
\(\dfrac{x+2}{-1} = \dfrac{y+3}{7} = \dfrac{z-3}{\frac{3}{2}}\)
Hence direction ratios of the given line are
\((-1,\, 7,\, \tfrac{3}{2})\)
Multiplying by 2 to remove fraction, we get
\((-2,\, 14,\, 3)\)
The required line is parallel to the given line.
\(\Rightarrow\) It has same direction ratios \( (-2,14,3) \).
Since it passes through \( (1,2,3) \),
Cartesian (symmetric) form of the required line is
\(\dfrac{x-1}{-2} = \dfrac{y-2}{14} = \dfrac{z-3}{3}\)
Vector form:
Position vector of point \( (1,2,3) \) is \( (\hat{i} + 2\hat{j} + 3\hat{k}) \)
Hence vector equation is
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-2\hat{i} + 14\hat{j} + 3\hat{k})\)
\(\therefore\) Required cartesian form is
\(\dfrac{x-1}{-2} = \dfrac{y-2}{14} = \dfrac{z-3}{3}\)
and vector form
\(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-2\hat{i} + 14\hat{j} + 3\hat{k})\) .
Solution:
Direction vector of first line is
\((2,-2,1)\)
Direction vector of second line is
\((1,2,2)\)
The required line is perpendicular to both lines.
\(\Rightarrow\) Its direction ratios are the cross product of \((2,-2,1)\) and \((1,2,2)\).
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix} \)
\(= \hat{i}((-2)(2) - (1)(2)) - \hat{j}((2)(2) - (1)(1)) + \hat{k}((2)(2) - (-2)(1))\)
\(= \hat{i}(-4 - 2) - \hat{j}(4 - 1) + \hat{k}(4 + 2)\)
\(= -6\hat{i} - 3\hat{j} + 6\hat{k}\)
Direction ratios \(= (-6,-3,6)\)
On dividing by 3, we get
\((-2,-1,2)\)
Since line passes through \( (2,-1,3) \),
Cartesian form:
\(\dfrac{x-2}{-2} = \dfrac{y+1}{-1} = \dfrac{z-3}{2}\)
Vector form:
Position vector of point \( (2,-1,3) \) is \( (2\hat{i} - \hat{j} + 3\hat{k}) \)
\(\therefore \vec r = (2\hat{i} - \hat{j} + 3\hat{k}) + t(-2\hat{i} - \hat{j} + 2\hat{k})\)
\(\therefore\) Required cartesian form is
\(\dfrac{x-2}{-2} = \dfrac{y+1}{-1} = \dfrac{z-3}{2}\)
and vector form
\(\vec r = (2\hat{i} - \hat{j} + 3\hat{k}) + t(-2\hat{i} - \hat{j} + 2\hat{k})\) .
Solution:
Direction ratios of first line are
\((3,-16,7)\)
Direction ratios of second line are
\((3,8,-5)\)
The required line is perpendicular to both lines.
\(\Rightarrow\) Its direction ratios are the cross product of \((3,-16,7)\) and \((3,8,-5)\).
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \)
\(= \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))\)
\(= \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)\)
\(= 24\hat{i} + 36\hat{j} + 72\hat{k}\)
Direction ratios \(= (24,36,72)\)
Dividing by 12, we get as DRs \((2,3,6)\)
Since it passes through \( (1,2,-4) \),
Position vector of the point is \( (\hat{i} + 2\hat{j} - 4\hat{k}) \).
Hence vector equation is
\(\vec r = (\hat{i} + 2\hat{j} - 4\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})\)
\(\therefore\) Required vector equation is
\(\vec r = (\hat{i} + 2\hat{j} - 4\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})\) .
Solution:
Given first line:
\(\dfrac{2x-1}{4} = \dfrac{3y+5}{2} = \dfrac{2-x}{3}\)
Rewrite in standard form.
\(\dfrac{x-\frac{1}{2}}{2} = \dfrac{y+\frac{5}{3}}{\frac{2}{3}} = \dfrac{x-2}{-3}\)
Direction ratios of first line are
\((2,\, \tfrac{2}{3},\, -3)\)
Multiplying by 3 to remove fraction, we get DRs as \((6,\,2,\,-9)\)
Second line:
\(\dfrac{x}{-3} = \dfrac{y}{2} = \dfrac{z}{5}\)
Direction ratios are \((-3,\,2,\,5)\)
The required line is perpendicular to both lines.
\(\Rightarrow\) Its direction ratios are the cross product of \((6,2,-9)\) and \((-3,2,5)\).
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 2 & -9 \\ -3 & 2 & 5 \end{vmatrix} \)
\(= \hat{i}(2\cdot5 - (-9)\cdot2) - \hat{j}(6\cdot5 - (-9)(-3)) + \hat{k}(6\cdot2 - 2(-3))\)
\(= \hat{i}(10 + 18) - \hat{j}(30 - 27) + \hat{k}(12 + 6)\)
\(= 28\hat{i} - 3\hat{j} + 18\hat{k}\)
Direction ratios \(= (28,-3,18)\)
Given point has position vector \( (-\hat{i} + 2\hat{j} + \hat{k}) \).
Hence vector equation is
\(\vec r = (-\hat{i} + 2\hat{j} + \hat{k}) + t(28\hat{i} - 3\hat{j} + 18\hat{k})\)
\(\therefore\) Required vector equation is
\(\vec r = (-\hat{i} + 2\hat{j} + \hat{k}) + t(28\hat{i} - 3\hat{j} + 18\hat{k})\) .
Solution:
Given line:
\(\dfrac{x-11}{10} = \dfrac{y+2}{-4} = \dfrac{z+8}{-11}\)
A point on the line is \(A(11,-2,-8)\)
Direction ratios of the line are \((10,-4,-11)\)
Let the foot of perpendicular be \(Q(x,y,z)\) on the given line.
Parametric form of the line is
\(x = 11 + 10t\)
\(y = -2 - 4t\)
\(z = -8 - 11t\)
So, \(Q(11+10t,\,-2-4t,\,-8-11t)\)
Given point \(P(2,-1,5)\).
Since \(PQ \perp\) given line,
\(\Rightarrow \overrightarrow{PQ} \cdot (10,-4,-11) = 0 \quad .... (i)\)
\(\overrightarrow{PQ} = (11+10t-2,\,-2-4t+1,\,-8-11t-5)\)
\(= (9+10t,\,-1-4t,\,-13-11t)\)
Now from equation \((i)\), we get
\((9+10t)(10) + (-1-4t)(-4) + (-13-11t)(-11) = 0\)
\(\Rightarrow 90 + 100t + 4 + 16t + 143 + 121t = 0\)
\(\Rightarrow 237 + 237t = 0\)
\(\Rightarrow 237(1+t) = 0\)
\(\Rightarrow t = -1\)
Substitute \(t=-1\) in coordinates of \(Q\):
\(x = 11 - 10 = 1\)
\(y = -2 + 4 = 2\)
\(z = -8 + 11 = 3\)
\(\therefore\) Foot of perpendicular is \((1,2,3)\).
Perpendicular distance:
\(PQ = \sqrt{(1-2)^2 + (2+1)^2 + (3-5)^2}\)
\(= \sqrt{(-1)^2 + 3^2 + (-2)^2}\)
\(= \sqrt{1 + 9 + 4}\)
\(= \sqrt{14}\)
\(\therefore\) Perpendicular distance \(= \) \(\sqrt{14}\).
Solution:
Given line:
\(\dfrac{x+2}{3} = \dfrac{y+1}{2} = \dfrac{z-3}{2}\)
Let \(\dfrac{x+2}{3} = \dfrac{y+1}{2} = \dfrac{z-3}{2} = t\). So,
\(\Rightarrow x = -2 + 3t\)
\(\Rightarrow y = -1 + 2t\)
\(\Rightarrow z = 3 + 2t\)
Let the required point on the line be
\(P(-2+3t,\,-1+2t,\,3+2t)\)
Given point \(A(1,2,3)\).
Distance \(AP = 3\sqrt{2}\).
\( AP^2 = [(-2+3t-1)]^2 + [(-1+2t-2)]^2 + [(3+2t-3)]^2 \)
\( = (-3+3t)^2 + (-3+2t)^2 + (2t)^2 \)
\( = 9(t-1)^2 + (2t-3)^2 + 4t^2 \)
\( = 9(t^2 - 2t + 1) + (4t^2 -12t +9) + 4t^2 \)
\( = 9t^2 -18t +9 + 4t^2 -12t +9 + 4t^2 \)
\( = 17t^2 -30t +18 \)
Since \(AP = 3\sqrt{2}\),
\( AP^2 = 18 \)
\( \Rightarrow 17t^2 -30t +18 = 18 \)
\( \Rightarrow 17t^2 -30t = 0 \)
\( \Rightarrow t(17t - 30) = 0 \)
\(\Rightarrow t = 0 \quad \text{or} \quad t = \dfrac{30}{17}\)
When \(t=0\):
\(P(-2,-1,3)\)
When \(t=\dfrac{30}{17}\):
\( x = -2 + \frac{90}{17} = \frac{56}{17} \)
\( y = -1 + \frac{60}{17} = \frac{43}{17} \)
\( z = 3 + \frac{60}{17} = \frac{111}{17} \)
\(\therefore\) Required points are
\((-2,-1,3)\)
and
\(\left(\dfrac{56}{17},\,\dfrac{43}{17},\,\dfrac{111}{17}\right)\) .
Solution:
Given line is
\(\vec r = -\hat{i} + 3\hat{j} + \hat{k} + \lambda(2\hat{i} + 3\hat{j} - \hat{k})\)
It can be written in cartesian form as
\(\dfrac{x+1}{2} = \dfrac{y-3}{3} = \dfrac{z-1}{-1} \quad ...(i)\)
Let \(Q(\alpha,\beta,\gamma)\) be the foot of perpendicular drawn from \(P(5,4,2)\) to line (i)
\(\therefore Q(\alpha,\beta,\gamma)\) lies on line (i)
\(\dfrac{\alpha+1}{2} = \dfrac{\beta-3}{3} = \dfrac{\gamma-1}{-1} = \lambda\)
\(\Rightarrow \alpha = 2\lambda -1\)
\(\beta = 3\lambda +3\)
\(\gamma = -\lambda +1\)
Now,
\(\overrightarrow{PQ} = (\alpha-5)\hat{i} + (\beta-4)\hat{j} + (\gamma-2)\hat{k}\)
Parallel vector of line (i) is
\(\vec b = 2\hat{i} + 3\hat{j} - \hat{k}\)
Obviously \( \overrightarrow{PQ} \perp \vec b \)
\(\therefore \overrightarrow{PQ} \cdot \vec b = 0\)
\(2(\alpha-5) + 3(\beta-4) - (\gamma-2) = 0\)
\(\Rightarrow 2\alpha -10 + 3\beta -12 -\gamma +2 = 0\)
\(\Rightarrow 2\alpha + 3\beta - \gamma -20 = 0\)
Putting values of \(\alpha,\beta,\gamma\),
\(2(2\lambda-1) + 3(3\lambda+3) - (-\lambda+1) -20 = 0\)
\(\Rightarrow 4\lambda -2 + 9\lambda +9 + \lambda -1 -20 = 0\)
\(\Rightarrow 14\lambda -14 = 0\)
\(\Rightarrow \lambda = 1\)
Hence coordinates of foot of perpendicular \(Q\) are
\((2\times1 -1,\; 3\times1 +3,\; -1+1)\)
\(= (1,6,0)\)
Length of perpendicular:
\( PQ = \sqrt{(5-1)^2 + (4-6)^2 + (2-0)^2} \)
\( = \sqrt{16 + 4 + 4} \)
\( = \sqrt{24} = 2\sqrt{6} \)
Also since \(Q\) is mid-point of \(PP'\)
\(1 = \dfrac{x_1 + 5}{2} \Rightarrow x_1 = -3\)
\(6 = \dfrac{y_1 + 4}{2} \Rightarrow y_1 = 8\)
\(0 = \dfrac{z_1 + 2}{2} \Rightarrow z_1 = -2\)
\(\therefore\) Required image is \((-3,8,-2)\).
Solution:
Given line is
\(\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}\)
Let \(Q(\alpha,\beta,\gamma)\) be the foot of perpendicular drawn from \(P(1,6,3)\) to the given line.
\(\therefore Q(\alpha,\beta,\gamma)\) lie on the given line
\(\dfrac{\alpha}{1} = \dfrac{\beta-1}{2} = \dfrac{\gamma-2}{3} = \lambda \; (say)\)
\(\Rightarrow \alpha = \lambda\)
\(\Rightarrow \beta = 1 + 2\lambda\)
\(\Rightarrow \gamma = 2 + 3\lambda\)
Now,
\(\overrightarrow{PQ} = (\alpha-1)\hat{i} + (\beta-6)\hat{j} + (\gamma-3)\hat{k}\)
Parallel vector of given line is
\(\vec b = \hat{i} + 2\hat{j} + 3\hat{k}\)
Obviously \( \overrightarrow{PQ} \perp \vec b \)
\(\therefore \overrightarrow{PQ} \cdot \vec b = 0\)
\((\alpha-1)(1) + (\beta-6)(2) + (\gamma-3)(3) = 0\)
\(\Rightarrow \alpha -1 + 2\beta -12 + 3\gamma -9 = 0\)
\(\Rightarrow \alpha + 2\beta + 3\gamma -22 = 0\)
Putting \(\alpha,\beta,\gamma\),
\(\lambda + 2(1+2\lambda) + 3(2+3\lambda) -22 = 0\)
\(\Rightarrow \lambda + 2 + 4\lambda + 6 + 9\lambda -22 = 0\)
\(\Rightarrow 14\lambda -14 = 0\)
\(\Rightarrow \lambda = 1\)
Hence coordinates of foot of perpendicular \(Q\) are
\((1,\;3,\;5)\)
Also since \(Q\) is mid-point of \(PP'\)
\(1 = \dfrac{x_1 + 1}{2} \Rightarrow x_1 = 1\)
\(3 = \dfrac{y_1 + 6}{2} \Rightarrow y_1 = 0\)
\(5 = \dfrac{z_1 + 3}{2} \Rightarrow z_1 = 7\)
\(\therefore\) Required image is \((1,0,7)\).
Equation of line joining \(P\) and \(P'\):
Direction ratios \(= (1-1,\;0-6,\;7-3)\)
\(= (0,-6,4)\)
Dividing by 2, DRs \(= (0,-3,2)\)
Hence equation is
\(\dfrac{x-1}{0} = \dfrac{y-6}{-3} = \dfrac{z-3}{2}\)
\(\therefore\) Required line is
\(\dfrac{y-6}{-3} = \dfrac{z-3}{2}, \quad x=1 \) .
Solution:
First line:
\(\vec r = (4\hat{i} - \hat{k}) + \lambda(2\hat{i} + 3\hat{k})\)
\(\Rightarrow x = 4 + 2\lambda\)
\(\Rightarrow y = 0\)
\(\Rightarrow z = -1 + 3\lambda\)
Second line:
\(\vec r = (\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - \hat{j})\)
\(\Rightarrow x = 1 + 3\mu\)
\(\Rightarrow y = 1 - \mu\)
\(\Rightarrow z = -1\)
At point of intersection, coordinates are equal.
From \(z\)-coordinates:
\(-1 + 3\lambda = -1\)
\(\Rightarrow 3\lambda = 0\)
\(\Rightarrow \lambda = 0\)
Then from first line:
\(x = 4,\; y = 0,\; z = -1\)
Now from second line:
\(4 = 1 + 3\mu\)
\(\Rightarrow 3\mu = 3\)
\(\Rightarrow \mu = 1\)
\(y = 1 - 1 = 0\) (It satisfied the coordinates of point intersection on first line)
\(\therefore\) Point of intersection is
\((4,0,-1)\).
Solution:
First line:
\(\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4} = \lambda\)
\(\Rightarrow x = 1 + 2\lambda\)
\(\Rightarrow y = -1 + 3\lambda\)
\(\Rightarrow z = 1 + 4\lambda\)
Second line:
\(\dfrac{x-3}{1} = \dfrac{y-K}{2} = \dfrac{z}{1} = \mu\)
\(\Rightarrow x = 3 + \mu\)
\(\Rightarrow y = K + 2\mu\)
\(\Rightarrow z = \mu\)
Since the lines intersect, their coordinates must be equal for some \(\lambda\) and \(\mu\).
Equating \(z\)-coordinates:
\(1 + 4\lambda = \mu \quad ...(1)\)
Equating \(x\)-coordinates:
\(1 + 2\lambda = 3 + \mu \quad ...(2)\)
Substitute (1) into (2):
\(1 + 2\lambda = 3 + (1 + 4\lambda)\)
\(\Rightarrow 1 + 2\lambda = 4 + 4\lambda\)
\(\Rightarrow -3 = 2\lambda\)
\(\Rightarrow \lambda = -\dfrac{3}{2}\)
From (1),
\(\mu = 1 + 4\left(-\dfrac{3}{2}\right)\)
\(\Rightarrow \mu = 1 - 6 = -5\)
Now equating \(y\)-coordinates:
\(-1 + 3\lambda = K + 2\mu\)
\(-1 + 3\left(-\dfrac{3}{2}\right) = K + 2(-5)\)
\(-1 - \dfrac{9}{2} = K - 10\)
\(-\dfrac{11}{2} = K - 10\)
\(K = -\dfrac{11}{2} + 10\)
\(K = \dfrac{9}{2}\)
\(\therefore\) Required value of \(K\) is \(\dfrac{9}{2}\).
Solution:
Equations of given lines
\(\vec r = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}\)
or \(\vec r = \hat{i} - 2\hat{j} + 3\hat{k} + t(-\hat{i} + \hat{j} - 2\hat{k})\)
and \(\vec r = (s+1)\hat{i} + (2s-1)\hat{j} + (2s+1)\hat{k}\)
or \(\vec r = \hat{i} - \hat{j} + \hat{k} + s(\hat{i} + 2\hat{j} + 2\hat{k})\)
Comparing the above equations with \( \vec r = \vec a_1 + \lambda \vec b_1 \) and \( \vec r = \vec a_2 + \mu \vec b_2 \),
\(\vec a_1 = \hat{i} - 2\hat{j} + 3\hat{k}, \quad \vec b_1 = -\hat{i} + \hat{j} - 2\hat{k}\)
\(\vec a_2 = \hat{i} - \hat{j} + \hat{k}, \quad \vec b_2 = \hat{i} + 2\hat{j} + 2\hat{k}\)
\(\therefore \vec a_2 - \vec a_1 = (\hat{i} - \hat{j} + \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})\)
\(= \hat{j} - 2\hat{k}\)
\(\vec b_1 \times \vec b_2 = (-\hat{i} + \hat{j} - 2\hat{k}) \times (\hat{i} + 2\hat{j} + 2\hat{k})\)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & 2 \end{vmatrix} \)
\(= (1\cdot2 - (-2)\cdot2)\hat{i} -((-1)\cdot2 - (-2)\cdot1)\hat{j} +((-1)\cdot2 - 1\cdot1)\hat{k}\)
\(= (2+4)\hat{i} -(-2+2)\hat{j} +(-2-1)\hat{k}\)
\(= 6\hat{i} - 0\hat{j} - 3\hat{k}\)
\(= 6\hat{i} - 3\hat{k}\)
\(|\vec b_1 \times \vec b_2| = \sqrt{6^2 + 0^2 + (-3)^2}\)
\(= \sqrt{36 + 9}\)
\(= \sqrt{45} = 3\sqrt{5}\)
\(\therefore\) Required minimum distance
\( d = \left| \dfrac{(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2)} {|\vec b_1 \times \vec b_2|} \right| \)
\( = \left| \dfrac{(\hat{j} - 2\hat{k})\cdot(6\hat{i} - 3\hat{k})} {3\sqrt{5}} \right| \)
\( = \left| \dfrac{0 + 6} {3\sqrt{5}} \right| \)
\( = \dfrac{6}{3\sqrt{5}} = \dfrac{2}{\sqrt{5}} = \dfrac{2\sqrt{5}}{5} \)
\(\therefore\) The minimum distance between the given lines is \(\dfrac{2\sqrt{5}}{5}\).
Solution:
Equations of given lines
\(\vec r = 4\hat{i} + 4\hat{j} + \hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})\)
and
\(\vec r = (\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i} + 2\hat{j} - 2\hat{k})\)
Comparing with \( \vec r = \vec a_1 + \lambda \vec b_1 \) and \( \vec r = \vec a_2 + \mu \vec b_2 \),
\(\vec a_1 = 4\hat{i} + 4\hat{j} + \hat{k}\)
\(\vec b_1 = \hat{i} + \hat{j} - \hat{k}\)
\(\vec a_2 = \hat{i} - \hat{j} - \hat{k}\)
\(\vec b_2 = \hat{i} + 2\hat{j} - 2\hat{k}\)
\(\therefore \vec a_2 - \vec a_1 = (\hat{i} - \hat{j} - \hat{k}) - (4\hat{i} + 4\hat{j} + \hat{k})\)
\(= -3\hat{i} -5\hat{j} -2\hat{k}\)
\(\vec b_1 \times \vec b_2 = (\hat{i} + \hat{j} - \hat{k}) \times (\hat{i} + 2\hat{j} - 2\hat{k})\)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 2 & -2 \end{vmatrix} \)
\(= (1\cdot(-2) - (-1)\cdot2)\hat{i} - (1\cdot(-2) - (-1)\cdot1)\hat{j} + (1\cdot2 - 1\cdot1)\hat{k}\)
\(= (-2+2)\hat{i} -(-2+1)\hat{j} +(2-1)\hat{k}\)
\(= 0\hat{i} + \hat{j} + \hat{k}\)
\(|\vec b_1 \times \vec b_2| = \sqrt{0^2 + 1^2 + 1^2}\)
\(= \sqrt{2}\)
\(\therefore\) Required minimum distance
\( d = \left| \dfrac{(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2)} {|\vec b_1 \times \vec b_2|} \right| \)
\( = \left| \dfrac{(-3\hat{i} -5\hat{j} -2\hat{k})\cdot(\hat{j} + \hat{k})} {\sqrt{2}} \right| \)
\( = \left| \dfrac{-5 -2} {\sqrt{2}} \right| \)
\( = \dfrac{7}{\sqrt{2}} = \dfrac{7\sqrt{2}}{2} \)
\(\therefore\) The minimum distance between the given lines is \(\dfrac{7\sqrt{2}}{2}\).
Solution:
First convert into vector form:
\(\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6} = \lambda\)
\(\Rightarrow x = 1 + 2\lambda\)
\(\Rightarrow y = 2 + 3\lambda\)
\(\Rightarrow z = -4 + 6\lambda\)
\(\therefore \vec r = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})\)
\(\dfrac{x-3}{4} = \dfrac{y-3}{6} = \dfrac{z+5}{12} = \mu\)
\(\Rightarrow x = 3 + 4\mu\)
\(\Rightarrow y = 3 + 6\mu\)
\(\Rightarrow z = -5 + 12\mu\)
\(\therefore \vec r = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})\)
Comparing with \( \vec r = \vec a_1 + \lambda \vec b_1 \) and \( \vec r = \vec a_2 + \mu \vec b_2 \),
\(\vec a_1 = \hat{i} + 2\hat{j} - 4\hat{k}\)
\(\vec b_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}\)
\(\vec a_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}\)
\(\vec b_2 = 4\hat{i} + 6\hat{j} + 12\hat{k}\)
\(\therefore \vec a_2 - \vec a_1 = (2\hat{i} + \hat{j} - \hat{k})\)
\(\vec b_1 \times \vec b_2 = (2\hat{i} + 3\hat{j} + 6\hat{k}) \times (4\hat{i} + 6\hat{j} + 12\hat{k})\)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 4 & 6 & 12 \end{vmatrix} \)
\(= 0\hat{i} - 0\hat{j} + 0\hat{k}\)
Since \( \vec b_1 \times \vec b_2 = 0 \), the lines are parallel.
For parallel lines, shortest distance is
\( d = \dfrac{|(\vec a_2 - \vec a_1) \times \vec b_1|} {|\vec b_1|} \)
\((\vec a_2 - \vec a_1) \times \vec b_1\)
\( = (2\hat{i} + \hat{j} - \hat{k}) \times (2\hat{i} + 3\hat{j} + 6\hat{k}) \)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} \)
\(= (1\cdot6 - (-1)\cdot3)\hat{i} - (2\cdot6 - (-1)\cdot2)\hat{j} + (2\cdot3 - 1\cdot2)\hat{k}\)
\(= (6+3)\hat{i} -(12+2)\hat{j} +(6-2)\hat{k}\)
\(= 9\hat{i} -14\hat{j} +4\hat{k}\)
\(|(\vec a_2 - \vec a_1) \times \vec b_1| = \sqrt{9^2 + (-14)^2 + 4^2}\)
\(= \sqrt{81 + 196 + 16}\)
\(= \sqrt{293}\)
\(|\vec b_1| = \sqrt{2^2 + 3^2 + 6^2}\)
\(= \sqrt{4 + 9 + 36}\)
\(= \sqrt{49} = 7\)
\(\therefore d = \dfrac{\sqrt{293}}{7}\)
\(\therefore\) Shortest distance = \( \dfrac{\sqrt{293}}{7} \)
Solution:
Equations of given lines
\(\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}\)
or \(\vec r = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})\)
and
\(\dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5}\)
or \(\vec r = 2\hat{i} + 4\hat{j} + 5\hat{k} + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})\)
Comparing with \( \vec r = \vec a_1 + \lambda \vec b_1 \) and \( \vec r = \vec a_2 + \mu \vec b_2 \),
\(\vec a_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec b_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}\)
\(\vec a_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}, \quad \vec b_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}\)
\(\therefore \vec a_2 - \vec a_1 = (2\hat{i} + 4\hat{j} + 5\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})\)
\(= \hat{i} + 2\hat{j} + 2\hat{k}\)
\(\vec b_1 \times \vec b_2\)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} \)
\(= (3\cdot5 - 4\cdot4)\hat{i} - (2\cdot5 - 4\cdot3)\hat{j} + (2\cdot4 - 3\cdot3)\hat{k}\)
\(= (15-16)\hat{i} - (10-12)\hat{j} + (8-9)\hat{k}\)
\(= -\hat{i} + 2\hat{j} - \hat{k}\)
\(|\vec b_1 \times \vec b_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2}\)
\(= \sqrt{1 + 4 + 1}\)
\(= \sqrt{6}\)
\(\therefore\) Required minimum distance
\( d = \left| \dfrac{(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2)} {|\vec b_1 \times \vec b_2|} \right| \)
\( = \left| \dfrac{(\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k})} {\sqrt{6}} \right| \)
\( = \left| \dfrac{-1 + 4 - 2} {\sqrt{6}} \right| \)
\( = \dfrac{1}{\sqrt{6}} \)
\(\therefore\) Minimum distance between the given lines is \(\dfrac{1}{\sqrt{6}}\).
Equation of line of shortest distance:
It is parallel to \( \vec b_1 \times \vec b_2 = (-1,2,-1)\).
Passing through point on first line when \( \lambda = 0 \) are \((1,2,3)\)
\(\therefore\) Required line is
\( \vec r = (1,2,3) + t(-1,2,-1) \)
or
\( \dfrac{x-1}{-1} = \dfrac{y-2}{2} = \dfrac{z-3}{-1} \)
Solution:
Given line:
\(\dfrac{x-6}{2} = \dfrac{y-7}{2} = \dfrac{z-7}{-3}\)
Let each ratio be \( \lambda \).
\(\Rightarrow x = 6 + 2\lambda\)
\(\Rightarrow y = 7 + 2\lambda\)
\(\Rightarrow z = 7 - 3\lambda\)
Thus vector form:
\(\vec r = (6\hat{i} + 7\hat{j} + 7\hat{k}) + \lambda(2\hat{i} + 2\hat{j} - 3\hat{k})\)
Let \(Q(\alpha,\beta,\gamma)\) be the foot of perpendicular from \(P(1,2,3)\).
\(\Rightarrow Q(6+2\lambda,\;7+2\lambda,\;7-3\lambda)\)
Direction vector of line is \((2,2,-3)\).
Since \(PQ \perp\) line,
\(\overrightarrow{PQ} \cdot (2,2,-3) = 0\)
\(\overrightarrow{PQ} = (6+2\lambda-1,\;7+2\lambda-2,\;7-3\lambda-3)\)
\(= (5+2\lambda,\;5+2\lambda,\;4-3\lambda)\)
Now,
\((5+2\lambda)2 + (5+2\lambda)2 + (4-3\lambda)(-3) = 0\)
\(\Rightarrow 10+4\lambda + 10+4\lambda -12+9\lambda = 0\)
\(\Rightarrow 8 + 17\lambda = 0\)
\(\Rightarrow \lambda = -\dfrac{8}{17}\)
Hence coordinates of foot \(Q\):
\( x = 6 + 2\left(-\dfrac{8}{17}\right) = \dfrac{86}{17} \)
\( y = 7 + 2\left(-\dfrac{8}{17}\right) = \dfrac{103}{17} \)
\( z = 7 - 3\left(-\dfrac{8}{17}\right) = \dfrac{143}{17} \)
\(\therefore\) Foot of perpendicular is
\(\left(\dfrac{86}{17}, \dfrac{103}{17}, \dfrac{143}{17}\right)\)
Perpendicular distance:
\( PQ = \sqrt{ \left(\dfrac{86}{17}-1\right)^2 + \left(\dfrac{103}{17}-2\right)^2 + \left(\dfrac{143}{17}-3\right)^2 } \)
\( = \sqrt{\left(\dfrac{69}{17}\right)^2 +\left(\dfrac{69}{17}\right)^2 +\left(\dfrac{92}{17}\right)^2} \)
\( = \dfrac{1}{17}\sqrt{69^2+69^2+92^2} \)
\( = \dfrac{1}{17}\sqrt{4761+4761+8464} \)
\( = \dfrac{1}{17}\sqrt{17986} \)
\(\therefore\) Perpendicular distance \(= \dfrac{\sqrt{17986}}{17}\).
Vector form of foot:
\( \vec r = \dfrac{86}{17}\hat{i} + \dfrac{103}{17}\hat{j} + \dfrac{143}{17}\hat{k} \)
Solution:
Given:
\( al + bm + cn = 0 \quad ...(1)\)
\( fmn + gnl + hlm = 0 \quad ...(2)\)
From (1),
\( n = -\left(\dfrac{al + bm}{c}\right) \quad ...(3)\)
Substituting this value of \(n\) in equation (2), we get
\((fm + gl)\left[-\dfrac{al + bm}{c}\right] + hlm = 0\)
\(\therefore - (aflm + bfm^2 + agl^2 + bglm) + chlm = 0\)
\(\therefore agl^2 + (af + bg - ch)lm + bfm^2 = 0 \quad ...(4)\)
Note that both \(l\) and \(m\) cannot be zero, because if \(l = m = 0\), then from (3), we get \(n = 0\), which is not possible as \(l^2 + m^2 + n^2 = 1\).
Let us take \(m \neq 0\).
Dividing equation (4) by \(m^2\), we get
\( ag\left(\dfrac{1}{m^2}\right) + (af + bg - ch)\left(\dfrac{1}{m}\right) + bf = 0 \quad ...(5) \)
This is quadratic equation in \( \left(\dfrac{1}{m}\right) \).
If \(l_1, m_1, n_1\) and \(l_2, m_2, n_2\) are the direction cosines of the two lines given by equations (1) and (2), then \( \dfrac{l_1}{m_1} \) and \( \dfrac{l_2}{m_2} \) are the roots of equation (5).
From the quadratic equation (5), we get
\( \text{Product of roots} = \dfrac{l_1}{m_1} \cdot \dfrac{l_2}{m_2} = \dfrac{bf}{ag} \)
\(\therefore \dfrac{l_1l_2}{m_1m_2} = \dfrac{f/a}{g/b}\)
\(\therefore \dfrac{l_1l_2}{f/a} = \dfrac{m_1m_2}{g/b}\)
Similarly, we can show that,
\(\dfrac{l_1l_2}{f/a} = \dfrac{n_1n_2}{h/c} = \lambda \quad ...(say)\)
\(\therefore l_1l_2 = \lambda \left(\dfrac{f}{a}\right),\; m_1m_2 = \lambda \left(\dfrac{g}{b}\right),\; n_1n_2 = \lambda \left(\dfrac{h}{c}\right)\)
Now, the lines are perpendicular if
\(l_1l_2 + m_1m_2 + n_1n_2 = 0\)
\(\Rightarrow \lambda\left(\dfrac{f}{a}\right) + \lambda\left(\dfrac{g}{b}\right) + \lambda\left(\dfrac{h}{c}\right) = 0\)
\(\Rightarrow \lambda\left( \dfrac{f}{a} + \dfrac{g}{b} + \dfrac{h}{c} \right) = 0\)
Since \( \lambda \neq 0 \),
\(\therefore \dfrac{f}{a} + \dfrac{g}{b} + \dfrac{h}{c} = 0\)
Hence Proved.
Solution:
Equations of given lines
\(\dfrac{x-1}{1} = \dfrac{y}{-1} = \dfrac{z}{2}\)
or \(\vec r = \hat{i} + t(\hat{i} - \hat{j} + 2\hat{k})\)
and
\(\dfrac{x+1}{2} = \dfrac{y}{2} = \dfrac{z-3}{\lambda}\)
or \(\vec r = (-\hat{i} + 3\hat{k}) + s(2\hat{i} + 2\hat{j} + \lambda\hat{k})\)
Comparing with \( \vec r = \vec a_1 + t\vec b_1 \) and \( \vec r = \vec a_2 + s\vec b_2 \),
\(\vec a_1 = \hat{i}, \quad \vec b_1 = \hat{i} - \hat{j} + 2\hat{k}\)
\(\vec a_2 = -\hat{i} + 3\hat{k}, \quad \vec b_2 = 2\hat{i} + 2\hat{j} + \lambda\hat{k}\)
\(\therefore \vec a_2 - \vec a_1 = (-\hat{i} + 3\hat{k}) - \hat{i}\)
\(= -2\hat{i} + 3\hat{k}\)
\(\vec b_1 \times \vec b_2\)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 2 & \lambda \end{vmatrix} \)
\(= (-\lambda - 4)\hat{i} -(\lambda - 4)\hat{j} + 4\hat{k}\)
\(|\vec b_1 \times \vec b_2| = \sqrt{(\lambda+4)^2 + (\lambda-4)^2 + 16}\)
\(= \sqrt{2\lambda^2 + 48}\)
\(= \sqrt{2(\lambda^2 + 24)}\)
\(\therefore\) Required minimum distance
\( d = \left| \dfrac{(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2)} {|\vec b_1 \times \vec b_2|} \right| \)
\( = \left| \dfrac{(-2\hat{i} + 3\hat{k}) \cdot [(-\lambda -4)\hat{i} -(\lambda -4)\hat{j} +4\hat{k}]} {\sqrt{2(\lambda^2+24)}} \right| \)
\( = \left| \dfrac{2(\lambda+4) + 12} {\sqrt{2(\lambda^2+24)}} \right| \)
\( = \left| \dfrac{2\lambda + 20} {\sqrt{2(\lambda^2+24)}} \right| \)
Given shortest distance \(= 1\)
\( \left| \dfrac{2\lambda + 20} {\sqrt{2(\lambda^2+24)}} \right| = 1 \)
\( \Rightarrow (2\lambda + 20)^2 = 2(\lambda^2 + 24) \)
\( \Rightarrow 4\lambda^2 + 80\lambda + 400 = 2\lambda^2 + 48 \)
\( \Rightarrow 2\lambda^2 + 80\lambda + 352 = 0 \)
\( \Rightarrow \lambda^2 + 40\lambda + 176 = 0 \)
\( \lambda = \dfrac{-40 \pm \sqrt{1600 - 704}}{2} \)
\( = \dfrac{-40 \pm \sqrt{896}}{2} \)
\( = -20 \pm 4\sqrt{14} \)
\(\therefore\) Required values of \( \lambda \) are
\(\lambda = -20 + 4\sqrt{14}\) or \(\lambda = -20 - 4\sqrt{14}\).
Solution:
Equation of line AB:
Direction vector \(\overrightarrow{AB} = (4-0,\;5+1,\;1+1)\)
\(= (4,6,2)\)
\(\therefore\) Line AB:
\(\vec r = (0,-1,-1) + t(4,6,2)\)
\(\Rightarrow x = 4t\)
\(\Rightarrow y = -1 + 6t\)
\(\Rightarrow z = -1 + 2t\)
Equation of line CD:
Direction vector \(\overrightarrow{CD} = (-4-3,\;4-9,\;4-4)\)
\(= (-7,-5,0)\)
\(\therefore\) Line CD:
\(\vec r = (3,9,4) + s(-7,-5,0)\)
\(\Rightarrow x = 3 - 7s\)
\(\Rightarrow y = 9 - 5s\)
\(\Rightarrow z = 4\)
For intersection, coordinates must be equal.
From \(z\)-coordinate:
\(-1 + 2t = 4\)
\(\Rightarrow 2t = 5\)
\(\Rightarrow t = \dfrac{5}{2}\)
Substitute in \(x\)-coordinate:
\(x = 4t = 10\)
Now equate with second line:
\(10 = 3 - 7s\)
\(\Rightarrow -7s = 7\)
\(\Rightarrow s = -1\)
Check \(y\)-coordinate:
\(y = -1 + 6t = -1 + 15 = 14\)
From second line:
\(y = 9 - 5(-1) = 14\) ✓
Thus both coordinates satisfy.
\(\therefore\) The two lines intersect at \((10,14,4)\).
Solution:
Given line:
\(\dfrac{x-11}{10} = \dfrac{y+2}{-4} = \dfrac{z+8}{-11}\)
Let each ratio be \( \lambda \).
\(\Rightarrow x = 11 + 10\lambda\)
\(\Rightarrow y = -2 - 4\lambda\)
\(\Rightarrow z = -8 - 11\lambda\)
Thus vector form:
\(\vec r = (11\hat{i} - 2\hat{j} - 8\hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 11\hat{k})\)
Let \(Q\) be the foot of perpendicular from \(P(2,-1,5)\).
\(Q(11+10\lambda,\; -2-4\lambda,\; -8-11\lambda)\)
Direction vector of line is \((10,-4,-11)\).
Since \(PQ \perp\) line,
\(\overrightarrow{PQ} \cdot (10,-4,-11) = 0\)
\(\overrightarrow{PQ} = (11+10\lambda-2,\; -2-4\lambda+1,\; -8-11\lambda-5)\)
\(= (9+10\lambda,\; -1-4\lambda,\; -13-11\lambda)\)
Now,
\((9+10\lambda)10 + (-1-4\lambda)(-4) + (-13-11\lambda)(-11) = 0\)
\(\Rightarrow 90+100\lambda + 4+16\lambda + 143+121\lambda = 0\)
\(\Rightarrow 237 + 237\lambda = 0\)
\(\Rightarrow \lambda = -1\)
Hence foot of perpendicular \(Q\):
\(x = 11 - 10 = 1\)
\(y = -2 + 4 = 2\)
\(z = -8 + 11 = 3\)
\(\therefore Q = (1,2,3)\)
Since \(Q\) is midpoint of \(PP'\),
\(1 = \dfrac{x_1 + 2}{2} \Rightarrow x_1 = 0\)
\(2 = \dfrac{y_1 -1}{2} \Rightarrow y_1 = 5\)
\(3 = \dfrac{z_1 + 5}{2} \Rightarrow z_1 = 1\)
\(\therefore\) Image point is \((0,5,1)\)
Equation of line joining \(P\) and \(P'\):
Direction ratios \(= (0-2,\;5+1,\;1-5)\)
\(= (-2,6,-4)\)
Dividing by 2, DRs \(= (-1,3,-2)\)
\(\therefore\) Required line is
\( \vec r = (2,-1,5) + t(-1,3,-2) \)
or
\( \dfrac{x-2}{-1} = \dfrac{y+1}{3} = \dfrac{z-5}{-2} \)
Solution:
Side PQ:
\(\overrightarrow{PQ} = (-1-0,\;6-7,\;6-10)\)
\(= (-1,-1,-4)\)
\(|PQ| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2}\)
\(= \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}\)
\(\therefore\) Direction cosines of \(PQ\):
\( \left( -\dfrac{1}{3\sqrt{2}},\; -\dfrac{1}{3\sqrt{2}},\; -\dfrac{4}{3\sqrt{2}} \right) \)
Side QR:
\(\overrightarrow{QR} = (-4+1,\;9-6,\;6-6)\)
\(= (-3,3,0)\)
\(|QR| = \sqrt{(-3)^2 + 3^2 + 0}\)
\(= \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}\)
\(\therefore\) Direction cosines of \(QR\):
\( \left( -\dfrac{1}{\sqrt{2}},\; \dfrac{1}{\sqrt{2}},\; 0 \right) \)
Side RP:
\(\overrightarrow{RP} = (0+4,\;7-9,\;10-6)\)
\(= (4,-2,4)\)
\(|RP| = \sqrt{4^2 + (-2)^2 + 4^2}\)
\(= \sqrt{16+4+16} = \sqrt{36} = 6\)
\(\therefore\) Direction cosines of \(RP\):
\( \left( \dfrac{2}{3},\; -\dfrac{1}{3},\; \dfrac{2}{3} \right) \)
\(\therefore\) Direction cosines of the three sides are
\(PQ:\left( -\dfrac{1}{3\sqrt{2}},\; -\dfrac{1}{3\sqrt{2}},\; -\dfrac{4}{3\sqrt{2}} \right)\)
\(QR:\left( -\dfrac{1}{\sqrt{2}},\; \dfrac{1}{\sqrt{2}},\; 0 \right)\)
\(RP:\left( \dfrac{2}{3},\; -\dfrac{1}{3},\; \dfrac{2}{3} \right)\)
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