CBSE Class 12 Mathematics Three Dimensional Geometry Important Questions Set A

Fill in the Blanks

Question. The distance between parallel planes \( 2x + y - 2z - 6 = 0 \) and \( 4x + 2y - 4z = 0 \) is _____________ units.
Answer: 2

Question. If \( p(1, 0, -3) \) is the foot of the perpendicular from the origin to the plane, then the Cartesian equation of the plane is _____________ .
Answer: \( x - 3z - 10 = 0 \)

Question. Vector equation of the line \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \) is ___________ .
Answer: \( \vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k}) \)

Question. If a line makes an angle of \( \frac{\pi}{4} \) with each of y and z–axis, then the angle which it makes with x – axis is _____________ .
Answer: \( \frac{\pi}{2} \)

Question. The cartesian equation of the plane \( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 2 \) is _____________ .
Answer: \( x + y - z = 2 \)

Solutions of Selected Fill in the Blanks

Question. Given parallel planes are \( 2x + y - 2z - 6 = 0 \) and \( 4x + 2y - 4z = 0 \).
Answer: Required distance \( D = \frac{|-6 - 0|}{\sqrt{(2)^2 + (1)^2 + (-2)^2}} = \frac{6}{3} = 2 \) units.

Question. Direction ratios of the normal to the plane are given by \( 1-0, 0-0, -3-0 \).
Answer: \( \Rightarrow 1, 0, -3 \). Equation of the plane be \( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \)
\( \Rightarrow 1 \cdot (x-1) + 0 \cdot (y-0) + (-3) \cdot (z+3) = 0 \)
\( \Rightarrow x - 1 - 3z - 9 = 0 \Rightarrow x - 3z - 10 = 0 \).

Question. Let the line makes angle \( \alpha \) with x–axis.
Answer: \( \therefore \cos^2 \alpha + \cos^2 \frac{\pi}{4} + \cos^2 \frac{\pi}{4} = 1 \)
\( \Rightarrow \cos^2 \alpha + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 \)
\( \Rightarrow \cos^2 \alpha + \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow \cos^2 \alpha = 0 \Rightarrow \alpha = \frac{\pi}{2} \).

Question. We have, \( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 2 \).
Answer: \( \Rightarrow (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 2 \Rightarrow x + y - z = 2 \).

Very Short Answer Questions

Question. If a line has direction ratios \( 2, -1, -2 \), then what are its direction cosines?
Answer: Direction cosines of line are \( \frac{2}{\sqrt{2^2 + (-1)^2 + (-2)^2}}, \frac{-1}{\sqrt{2^2 + (-1)^2 + (-2)^2}}, \frac{-2}{\sqrt{2^2 + (-1)^2 + (-2)^2}} \)
i.e., \( \frac{2}{3}, \frac{-1}{3}, \frac{-2}{3} \).

Question. Find the co-ordinate of the point, where the line \( \frac{x+2}{1} = \frac{y-5}{3} = \frac{z+1}{5} \) cuts the yz-plane.
Answer: Let the required point be \( (\alpha, \beta, \gamma) \) where given line cuts yz-plane.
\( \therefore \frac{\alpha+2}{1} = \frac{\beta-5}{3} = \frac{\gamma+1}{5} = k \text{ (say)} \)
Since point lies in yz-plane, \( \alpha = 0 \).
\( \Rightarrow 0 + 2 = k \Rightarrow k = 2 \)
So, \( \beta = 5 + 3k = 5 + 3(2) = 11 \) and \( \gamma = -1 + 5k = -1 + 5(2) = 9 \).
Required point is (0, 11, 9).

Question. Write the direction cosine of a line equally inclined to the three coordinate axes.
Answer: Any line equally inclined will have direction cosines \( l, l, l \).
\( \therefore l^2 + l^2 + l^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{3}} \).
Direction cosines are \( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \).

Question. Write the distance of the following plane from the origin. \( 2x - y + 2z + 1 = 0 \)
Answer: Distance from origin \( = \left| \frac{2(0) - 1(0) + 2(0) + 1}{\sqrt{2^2 + (-1)^2 + 2^2}} \right| = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{3} \).

Question. Find the acute angle between the planes \( \vec{r} \cdot (\hat{i} - 2\hat{j} - 2\hat{k}) = 1 \) and \( \vec{r} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 0 \).
Answer: \( \cos \theta = \frac{|(1)(3) + (-2)(-6) + (-2)(2)|}{\sqrt{1^2 + (-2)^2 + (-2)^2} \sqrt{3^2 + (-6)^2 + 2^2}} = \frac{|3 + 12 - 4|}{\sqrt{9}\sqrt{49}} = \frac{11}{3 \times 7} = \frac{11}{21} \).
\( \theta = \cos^{-1}\left(\frac{11}{21}\right) \).

Question. Write the direction cosines of a line parallel to z-axis.
Answer: The angles made are \( 90^\circ, 90^\circ, 0^\circ \).
Direction cosines are \( \cos 90^\circ, \cos 90^\circ, \cos 0^\circ \) i.e., \( 0, 0, 1 \).

Question. Write the cartesian equation of a plane, bisecting the line segment joining the points A(2, 3, 5) and B(4, 5, 7) at right angles.
Answer: Midpoint \( = \left(\frac{2+4}{2}, \frac{3+5}{2}, \frac{5+7}{2}\right) = (3, 4, 6) \).
Direction ratios of normal \( = (4-2, 5-3, 7-5) = (2, 2, 2) \).
Equation: \( 2(x-3) + 2(y-4) + 2(z-6) = 0 \Rightarrow 2x + 2y + 2z = 26 \) or \( x + y + z = 13 \).

Question. Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 \).
Answer: Normal of required plane \( = \hat{i} + \hat{j} + \hat{k} \).
Vector equation: \( (\vec{r} - (a\hat{i} + b\hat{j} + c\hat{k})) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0 \).

Question. Find the distance between the planes \( \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) - 4 = 0 \) and \( \vec{r} \cdot (6\hat{i} - 9\hat{j} + 18\hat{k}) + 30 = 0 \).
Answer: Second plane can be written as \( \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) + 10 = 0 \).
Distance \( = \frac{|10 - (-4)|}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{14}{\sqrt{4 + 9 + 36}} = \frac{14}{7} = 2 \) units.

Question. Write the equation of a plane which is at a distance of \( 5\sqrt{3} \) units from origin and the normal to which is equally inclined to coordinate axes.
Answer: Unit normal vector \( \hat{n} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \).
Equation: \( \vec{r} \cdot \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) = 5\sqrt{3} \Rightarrow x + y + z = 15 \).

Question. If a line makes angles \( 90^\circ \) and \( 60^\circ \) respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Answer: \( \cos^2 90^\circ + \cos^2 60^\circ + \cos^2 \theta = 1 \Rightarrow 0 + \frac{1}{4} + \cos^2 \theta = 1 \Rightarrow \cos^2 \theta = \frac{3}{4} \).
\( \cos \theta = \pm \frac{\sqrt{3}}{2} \Rightarrow \theta = 30^\circ \text{ or } 150^\circ \).

Question. Find the distance between the planes \( 2x - y + 2z = 5 \) and \( 5x - 2.5y + 5z = 20 \).
Answer: Dividing second plane by 2.5: \( 2x - y + 2z = 8 \).
Distance \( = \frac{|8 - 5|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{3}{3} = 1 \) unit.

Question. Find the equation of a plane that cuts the coordinates axes at (a, 0, 0), (0, b, 0) and (0, 0, c).
Answer: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).

Question. Find the angle between the line \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(3\hat{i} - \hat{j} + 2\hat{k}) \) and the plane \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 3 \).
Answer: \( \sin \theta = \frac{|(3)(1) + (-1)(1) + (2)(1)|}{\sqrt{3^2 + (-1)^2 + 2^2} \sqrt{1^2 + 1^2 + 1^2}} = \frac{|3-1+2|}{\sqrt{14}\sqrt{3}} = \frac{4}{\sqrt{42}} \).
\( \theta = \sin^{-1}\left(\frac{4}{\sqrt{42}}\right) \).

Short Answer Questions-I

Question. Find the points of intersection of the line \( \vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k}) \) and the plane \( \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \).
Answer: General point on line: \( (3\lambda + 2, 4\lambda - 1, 2\lambda + 2) \).
Substituting in plane: \( (3\lambda + 2) - (4\lambda - 1) + (2\lambda + 2) = 5 \)
\( \lambda + 5 = 5 \Rightarrow \lambda = 0 \).
Point is (2, –1, 2).

Question. If the x-coordinate of a point P on the join of Q(2, 2, 1) and R(5, 1, –2) is 4, then find its z-coordinate.
Answer: Let ratio be \( \lambda : 1 \). \( x = \frac{5\lambda + 2}{\lambda + 1} = 4 \Rightarrow 5\lambda + 2 = 4\lambda + 4 \Rightarrow \lambda = 2 \).
\( z = \frac{-2\lambda + 1}{\lambda + 1} = \frac{-2(2) + 1}{2 + 1} = \frac{-3}{3} = -1 \).

Question. Show that the points \( (\hat{i} - \hat{j} + 3\hat{k}) \) and \( 3(\hat{i} + \hat{j} + \hat{k}) \) are equidistant from the plane \( \vec{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0 \) and lies on opposite side of it.
Answer: Mid-point is \( 2\hat{i} + \hat{j} + 3\hat{k} \). Substituting in plane:
\( (2)(5) + (1)(2) + (3)(-7) + 9 = 10 + 2 - 21 + 9 = 0 \).
Since midpoint lies on the plane, the points are equidistant and on opposite sides.

Question. If the plane \( ax + by = 0 \) is rotated about its line of intersection with the plane \( z = 0 \) through an angle \( \alpha \), then prove that the equation of the plane in its new position is \( ax + by \pm z\sqrt{a^2 + b^2} \tan \alpha = 0 \).
Answer: Let new plane be \( ax + by + kz = 0 \). Angle between \( ax + by + 0z = 0 \) and \( ax + by + kz = 0 \) is \( \alpha \).
\( \cos \alpha = \frac{a^2 + b^2}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + k^2}} \Rightarrow \cos^2 \alpha = \frac{a^2 + b^2}{a^2 + b^2 + k^2} \)
\( k^2 = (a^2 + b^2) \tan^2 \alpha \Rightarrow k = \pm \sqrt{a^2 + b^2} \tan \alpha \).
Equation: \( ax + by \pm z\sqrt{a^2 + b^2} \tan \alpha = 0 \).

Question. Find the co-ordinates of the point where the line through (– 1, 1, – 8) and (5, – 2, 10) crosses the ZX - plane.
Answer: Equation of line: \( \frac{x+1}{6} = \frac{y-1}{-3} = \frac{z+8}{18} \).
In ZX-plane, \( y = 0 \). \( \frac{x+1}{6} = \frac{0-1}{-3} \Rightarrow x+1 = 2 \Rightarrow x = 1 \).
\( \frac{0-1}{-3} = \frac{z+8}{18} \Rightarrow z+8 = 6 \Rightarrow z = -2 \).
Point is (1, 0, – 2).

Short Answer Questions-II

Question. Find the shortest distance between the lines whose vector equations are:
\( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}) \).

Answer: Comparing the given equations with equations \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \).
We get \( \vec{a}_1 = \hat{i} + \hat{j}, \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}, \vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k} \)
Therefore, \( \vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{k}) \) and
\( \vec{b}_1 \times \vec{b}_2 = (2\hat{i} - \hat{j} + \hat{k}) \times (3\hat{i} - 5\hat{j} + 2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = 3\hat{i} - \hat{j} - 7\hat{k} \)
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{9 + 1 + 49} = \sqrt{59} \)
Hence, the shortest distance between the given lines is given by
\[ d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{3 - 0 + 7}{\sqrt{59}} \right| = \frac{10}{\sqrt{59}} \text{ units.} \]

Question. A line passes through (2, –1, 3) and is perpendicular to the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} - 3\hat{j} - 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \). Obtain its equation in vector and cartesian form.

Answer: Let \( \vec{b} \) be parallel vector of required line.
\( \Rightarrow \vec{b} \) is perpendicular to both given line.
\( \Rightarrow \vec{b} = (2\hat{i} - 2\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 2\hat{k}) \)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix} = (-4 - 2)\hat{i} - (4 - 1)\hat{j} + (4 + 2)\hat{k} = -6\hat{i} - 3\hat{j} + 6\hat{k} \).
Hence, the equation of line in vector form is
\( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(-6\hat{i} - 3\hat{j} + 6\hat{k}) \text{ or } \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \mu(2\hat{i} + \hat{j} - 2\hat{k}) \) [where \( \mu = -3\lambda \)]
Equation in cartesian form is
\( \frac{x - 2}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \)

Question. Find the equation of planes passing through the intersection of the planes \( \vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12 = 0 \) and \( \vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k}) = 0 \) and are at a unit distance from the origin.

Answer: We are given planes:
\( \vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12 = 0 \) ...(i)
\( \vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k}) = 0 \) ...(ii)
So equation of the required plane can be written as:
\( (\vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12) + \lambda(\vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k})) = 0 \)
\( \Rightarrow \vec{r} \cdot [(2 + 3\lambda)\hat{i} + (6 - \lambda)\hat{j} + 4\lambda\hat{k}] + 12 = 0 \) ...(iii)
In cartesian form
\( (2 + 3\lambda)x + (6 - \lambda)y + 4\lambda z + 12 = 0 \) ...(iv)
Since direction ratios of the normal to the plane are \( (2 + 3\lambda), (6 - \lambda), 4\lambda \); the direction cosines of it are:
\( \frac{2 + 3\lambda}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}}, \frac{6 - \lambda}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}}, \frac{4\lambda}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}} \)
So the distance of the plane from the origin is \( \frac{12}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}} \)
We are given that distance from origin is unity
\( \therefore \frac{12}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}} = 1 \)
\( \Rightarrow \frac{144}{4 + 9\lambda^2 + 12\lambda + 36 - 12\lambda + \lambda^2 + 16\lambda^2} = 1 \Rightarrow 144 = 26\lambda^2 + 40 \) (Squaring both sides)
\( \Rightarrow 26\lambda^2 = 104 \Rightarrow \lambda^2 = \frac{104}{26} = 4 \)
\( \Rightarrow \lambda = \pm 2 \)
\( \therefore \) Required equation of the plane is \( 8x + 4y + 8z + 12 = 0 \).
In vector form
\( \vec{r} \cdot (8\hat{i} + 4\hat{j} + 8\hat{k}) + 12 = 0 \)

Question. Find the vector equation of the plane determined by the points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6). Hence, find the distance of the plane, thus obtained from the origin.

Answer: Required equation of plane is given by:
\( \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 5 - 3 & 2 + 1 & 4 - 2 \\ -1 - 3 & -1 + 1 & 6 - 2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} = 0 \)
\( \Rightarrow (x - 3)12 - (y + 1)16 + (z - 2)12 = 0 \)
\( \Rightarrow 12x - 16y + 12z - 36 - 16 - 24 = 0 \Rightarrow 12x - 16y + 12z - 76 = 0 \Rightarrow 3x - 4y + 3z - 19 = 0 \)
Vector form:
\( \vec{r} \cdot (3\hat{i} - 4\hat{j} + 3\hat{k}) = 19 \). ...(i)
Distance of plane (i) from origin
\( \frac{\vec{r} \cdot (3\hat{i} - 4\hat{j} + 3\hat{k})}{\sqrt{9 + 16 + 9}} = \frac{19}{\sqrt{34}} \Rightarrow |\vec{r} \cdot \hat{n} = d| \)
Therefore distance of plane from origin is \( \frac{19}{\sqrt{34}} \) units.

Question. Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \( \frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3} \) and \( \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \).

Answer: Let the cartesian equation of the line passing through (2, 1, 3) be
\( \frac{x - 2}{a} = \frac{y - 1}{b} = \frac{z - 3}{c} \) …(i)
Since, line (i) is perpendicular to given line
\( \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \) …(ii)
and \( \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \) …(iii)
\( \therefore a + 2b + 3c = 0 \) …(iv)
\( -3a + 2b + 5c = 0 \) …(v)
From equation (iv) and (v), we get
\( \frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6} \Rightarrow \frac{a}{4} = \frac{b}{-14} = \frac{c}{8} = \lambda \text{ (say)} \)
\( \Rightarrow a = 4\lambda, b = -14\lambda, c = 8\lambda \)
Putting the value of \( a, b \) and \( c \) in (i), we get
\( \frac{x - 2}{4\lambda} = \frac{y - 1}{-14\lambda} = \frac{z - 3}{8\lambda} \Rightarrow \frac{x - 2}{4} = \frac{y - 1}{-14} = \frac{z - 3}{8} \)
\( \Rightarrow \frac{x - 2}{2} = \frac{y - 1}{-7} = \frac{z - 3}{4} \), which is the cartesian form.
The vector form is \( \vec{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 4\hat{k}) \).

Question. Find the distance of the point P (6, 5, 9) from the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6).

Answer: Plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6) is
\( \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 5 - 3 & 2 + 1 & 4 - 2 \\ -1 - 3 & -1 + 1 & 6 - 2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} = 0 \)
\( (x - 3) \begin{vmatrix} 3 & 2 \\ 0 & 4 \end{vmatrix} - (y + 1) \begin{vmatrix} 2 & 2 \\ -4 & 4 \end{vmatrix} + (z - 2) \begin{vmatrix} 2 & 3 \\ -4 & 0 \end{vmatrix} = 0 \)
\( \Rightarrow 12x - 36 - 16y - 16 + 12z - 24 = 0 \Rightarrow 3x - 4y + 3z - 19 = 0 \)
Distance of this plane from point P ( 6, 5, 9) is
\( \frac{|3(6) - 4(5) + 3(9) - 19|}{\sqrt{(3)^2 + (-4)^2 + (3)^2}} = \frac{|18 - 20 + 27 - 19|}{\sqrt{9 + 16 + 9}} = \frac{6}{\sqrt{34}} \text{ units.} \)

Question. Show that the lines \( \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} \) and \( \frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5} \) intersect. Also find their point of intersection.

Answer: Given lines are
\( \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} \) …(i)
\( \frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5} \) …(ii)
Let two lines (i) and (ii) intersect at a point \( P(\alpha, \beta, \gamma) \).
\( \Rightarrow (\alpha, \beta, \gamma) \) satisfy line (i)
\( \Rightarrow \frac{\alpha+1}{3} = \frac{\beta+3}{5} = \frac{\gamma+5}{7} = \lambda \text{ (say)} \)
\( \Rightarrow \alpha = 3\lambda - 1, \beta = 5\lambda - 3, \gamma = 7\lambda - 5 \), …(iii)
Again \( (\alpha, \beta, \gamma) \) also lie on (ii), we get
\( \frac{3\lambda - 1 - 2}{1} = \frac{5\lambda - 3 - 4}{3} = \frac{7\lambda - 5 - 6}{5} \Rightarrow \frac{3\lambda - 3}{1} = \frac{5\lambda - 7}{3} = \frac{7\lambda - 11}{5} \)
I II III
From I and II:
\( \frac{3\lambda - 3}{1} = \frac{5\lambda - 7}{3} \Rightarrow 9\lambda - 9 = 5\lambda - 7 \Rightarrow 4\lambda = 2 \Rightarrow \lambda = \frac{1}{2} \)
From II and III:
\( \frac{5\lambda - 7}{3} = \frac{7\lambda - 11}{5} \Rightarrow 25\lambda - 35 = 21\lambda - 33 \Rightarrow 4\lambda = 2 \Rightarrow \lambda = \frac{1}{2} \)
Since, the value of \( \lambda \) in both the cases is same
\( \Rightarrow \) Both lines intersect each other at a point.
\( \therefore \) Intersecting point \( = (\alpha, \beta, \gamma) = \left( \frac{3}{2} - 1, \frac{5}{2} - 3, \frac{7}{2} - 5 \right) \) [From (iii)]
\( = \left( \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2} \right) \)

Question. Find the vector and cartesian equations of the line passing through the point (1, 2, –4) and perpendicular to the two lines \( \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \) and \( \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \).
OR
Find the equation of a line passing through the point (1, 2, –4) and perpendicular to two lines \( \vec{r} = (8\hat{i} - 19\hat{j} + 10\hat{k}) + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k}) \) and \( \vec{r} = (15\hat{i} + 29\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) \).

Answer: Let the cartesian equation of line passing through (1, 2, – 4) be
\( \frac{x-1}{a} = \frac{y-2}{b} = \frac{z+4}{c} \) ...(i)
Given lines are
\( \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \) ...(ii)
\( \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \) ...(iii)
Obviously parallel vectors \( \vec{b}_1, \vec{b}_2 \) and \( \vec{b}_3 \) of (i), (ii) and (iii) respectively are given as
\( \vec{b}_1 = a\hat{i} + b\hat{j} + c\hat{k}, \vec{b}_2 = 3\hat{i} - 16\hat{j} + 7\hat{k}, \vec{b}_3 = 3\hat{i} + 8\hat{j} - 5\hat{k} \)
According to question
(i) \( \perp \) (ii) \( \Rightarrow \vec{b}_1 \perp \vec{b}_2 \Rightarrow \vec{b}_1 \cdot \vec{b}_2 = 0 \)
(i) \( \perp \) (iii) \( \Rightarrow \vec{b}_1 \perp \vec{b}_3 \Rightarrow \vec{b}_1 \cdot \vec{b}_3 = 0 \)
Hence, \( 3a - 16b + 7c = 0 \) ...(iv)
and \( 3a + 8b - 5c = 0 \) ...(v)
From equation (iv) and (v), we get
\( \frac{a}{80-56} = \frac{b}{21+15} = \frac{c}{24+48} \Rightarrow \frac{a}{24} = \frac{b}{36} = \frac{c}{72} \Rightarrow \frac{a}{2} = \frac{b}{3} = \frac{c}{6} = \lambda \text{ (say)} \)
\( \Rightarrow a= 2\lambda, b = 3\lambda, c = 6\lambda \)
Putting the value of \( a, b, c \) in (i), we get the required cartesian equation of line as
\( \frac{x-1}{2\lambda} = \frac{y-2}{3\lambda} = \frac{z+4}{6\lambda} \Rightarrow \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} \)
Hence, vector equation is
\( \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \)

Question. Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line \( 5x - 25 = 14 - 7y = 35z \).

Answer: Given line is \( 5x - 25 = 14 - 7y = 35z \)
\( \Rightarrow \frac{x-5}{1/5} = \frac{y-2}{-1/7} = \frac{z-0}{1/35} \Rightarrow \frac{x-5}{7} = \frac{y-2}{-5} = \frac{z-0}{1} \) ... (i)
Hence, parallel vector of given line i.e., \( \vec{b} = 7\hat{i} - 5\hat{j} + \hat{k} \)
Since required line is parallel to given line (i)
\( \Rightarrow \vec{b} = 7\hat{i} - 5\hat{j} + \hat{k} \) will also be parallel vector of required line which passes through A(1, 2, –1).
Therefore, required vector equation of line is
\( \vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k}) \)

Question. A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2} \).

Answer: Let the given variable plane meets X, Y and Z axes at A(a, 0, 0), B(0, b, 0), C(0, 0, c).
Therefore the equation of given plane is given by \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) … (i)
Let \( (\alpha, \beta, \gamma) \) be the coordinates of the centroid of triangle ABC. Then
\( \alpha = \frac{a + 0 + 0}{3} = \frac{a}{3} \Rightarrow a = 3\alpha; \beta = \frac{0 + b + 0}{3} = \frac{b}{3} \Rightarrow b = 3\beta; \gamma = \frac{0 + 0 + c}{3} = \frac{c}{3} \Rightarrow c = 3\gamma \)
\( \because \) 3p is the distance from origin to the plane (i)
\( \Rightarrow 3p = \left| \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{\sqrt{\left( \frac{1}{a} \right)^2 + \left( \frac{1}{b} \right)^2 + \left( \frac{1}{c} \right)^2}} \right| \Rightarrow \sqrt{\left( \frac{1}{a} \right)^2 + \left( \frac{1}{b} \right)^2 + \left( \frac{1}{c} \right)^2} = \frac{1}{3p} \)
Squaring both sides, we have
\( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9p^2} \Rightarrow \frac{1}{9\alpha^2} + \frac{1}{9\beta^2} + \frac{1}{9\gamma^2} = \frac{1}{9p^2} \) [Putting value of \( a = 3\alpha, b = 3\beta, c = 3\gamma \)]
\( \Rightarrow \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{1}{p^2} \)
Therefore, locus of \( (\alpha, \beta, \gamma) \) is \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2} \). Hence proved.