Three Dimensional Geometry JEE Mathematics Worksheets Set 01

Subjective Questions

Question. Show that points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) form an isosceles right angled triangle.
Answer: Let \( A (0, 7, 10) ; B(-1, 6, 6) ; C (-4, 9, 6) \)
\( AB = \sqrt{1 + 1 + 16} = \sqrt{18} \)
\( AC = \sqrt{16 + 4 + 16} = 6 \)
\( BC = \sqrt{9 + 9 + 0} = \sqrt{18} \)
Since \( AB = BC \) and \( AB^2 + BC^2 = 18 + 18 = 36 = AC^2 \), it forms an isosceles right angled triangle.

Question. Prove that the tetrahedron with vertices at the points (0, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0) is a regular tetrahedron. Find also the co-ordinates of its centroid.
Answer: Let the vertices be \( O(0,0,0), A(0,1,1), B(1,0,1), C(1,1,0) \).
The lengths of the edges are:
\( OA = \sqrt{0^2+1^2+1^2} = \sqrt{2} \)
\( OB = \sqrt{1^2+0^2+1^2} = \sqrt{2} \)
\( OC = \sqrt{1^2+1^2+0^2} = \sqrt{2} \)
\( AB = \sqrt{(1-0)^2+(0-1)^2+(1-1)^2} = \sqrt{2} \)
\( BC = \sqrt{(1-1)^2+(1-0)^2+(0-1)^2} = \sqrt{2} \)
\( CA = \sqrt{(0-1)^2+(1-1)^2+(1-0)^2} = \sqrt{2} \)
Since all edges are equal to \( \sqrt{2} \), it is a regular tetrahedron.
Centroid \( G = \left( \frac{0+0+1+1}{4}, \frac{0+1+0+1}{4}, \frac{0+1+1+0}{4} \right) = \left( \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right) \).

Question. Find the coordinates of the point equidistant from the point (a, 0, 0), (0, b, 0), (0, 0, c) and (0, 0, 0).
Answer: The point equidistant from the vertices of a tetrahedron is its circumcenter. For a tetrahedron with vertices at the origin and on the axes, the point is \( \left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right) \).

Question. Find the ratio in which the line joining the points (3, 5, -7) and (-2, 1, 8) is divided by the y-z plane. Find also the point of intersection on the plane and the line.
Answer: Let the ratio be \( \alpha : 1 \).
Point \( P = \left[ \frac{-2\alpha + 3}{\alpha + 1}, \frac{\alpha + 5}{\alpha + 1}, \frac{8\alpha - 7}{\alpha + 1} \right] \).
On y-z plane, \( x = 0 \).
\( \implies \frac{-2\alpha + 3}{\alpha + 1} = 0 \)
\( \implies \alpha = 3/2 \).
Ratio is 3 : 2.
Substituting \( \alpha \), the point \( P \) is \( \left( 0, \frac{13}{5}, 2 \right) \).

Question. What are the direction cosines of a line that passes through the points P(6, -7, -1) and Q(2, -3, 1) and is so directed that it makes an acute angle \( \alpha \) with the positive direction of x-axis.
Answer: \( \vec{QP} = (6-2, -7+3, -1-1) = (4, -4, -2) \).
Since it makes an acute angle with the positive x-axis, the direction ratios are \( (4, -4, -2) \) or \( 2(2, -2, -1) \).
Direction cosines = \( \left( \frac{2}{\sqrt{2^2+(-2)^2+(-1)^2}}, \frac{-2}{3}, \frac{-1}{3} \right) = \left( \frac{2}{3}, \frac{-2}{3}, \frac{-1}{3} \right) \).

Question. Find the angle between the lines whose direction cosines are given by \( \ell + m + n = 0 \) and \( \ell^2 + m^2 = n^2 \).
Answer: \( \ell + m + n = 0 \implies \ell = -(m+n) \).
Putting in \( \ell^2 + m^2 = n^2 \):
\( (m+n)^2 + m^2 = n^2 \)
\( \implies m^2 + n^2 + 2mn + m^2 = n^2 \)
\( \implies 2m^2 + 2mn = 0 \implies 2m(m+n) = 0 \).
Case (i): \( m = 0 \). Then \( \ell = -n \). Direction ratios are \( (-1, 0, 1) \).
Case (ii): \( m = -n \). Then \( \ell = 0 \). Direction ratios are \( (0, -1, 1) \).
Angle \( \cos \theta = \frac{|(-1)(0) + (0)(-1) + (1)(1)|}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \).
\( \implies \theta = 60^\circ \).

Question. Show that the foot of the perpendicular from the origin to the join of A(-9, 4, 5) and B(11, 0, -1) is the mid point of AB.
Answer: Midpoint of AB is \( C = \left( \frac{-9+11}{2}, \frac{4+0}{2}, \frac{5-1}{2} \right) = (1, 2, 2) \).
Direction ratios of OC are \( (1, 2, 2) \).
Direction ratios of AB are \( (11 - (-9), 0 - 4, -1 - 5) = (20, -4, -6) \).
Now, \( (1)(20) + (2)(-4) + (2)(-6) = 20 - 8 - 12 = 0 \).
Since \( OC \perp AB \), the midpoint \( C \) is the foot of the perpendicular from the origin.

Question. P and Q are the points (-1, 2, 1) and (4, 3, 5). Find the projection of PQ on a line which makes angles of 120 and 135 with y and z axes respectively and an acute angle with x-axis.
Answer: \( \vec{PQ} = (4 - (-1), 3 - 2, 5 - 1) = (5, 1, 4) \).
Let direction cosines of line be \( (\ell, m, n) \).
\( m = \cos 120^\circ = -1/2 \), \( n = \cos 135^\circ = -1/\sqrt{2} \).
\( \ell^2 + m^2 + n^2 = 1 \implies \ell^2 + 1/4 + 1/2 = 1 \implies \ell^2 = 1/4 \).
Since \( \alpha \) is acute, \( \ell = 1/2 \).
Direction cosines are \( \left( \frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}} \right) \).
Projection \( = |5(1/2) + 1(-1/2) + 4(-1/\sqrt{2})| = |2.5 - 0.5 - 2\sqrt{2}| = |2 - 2\sqrt{2}| = 2\sqrt{2} - 2 \).

Question. Find the equation of the planes passing through points (1, 0, 0) and (0, 1, 0) and making an angle of \( 0.25 \pi \) radians with plane \( x + y - 3 = 0 \).
Answer: Let the plane be \( Ax + By + Cz + 1 = 0 \).
Passes through (1, 0, 0) \( \implies A = -1 \).
Passes through (0, 1, 0) \( \implies B = -1 \).
Angle with \( x + y - 3 = 0 \) is \( \pi/4 \).
\( \cos(\pi/4) = \frac{1}{\sqrt{2}} = \frac{|(-1)(1) + (-1)(1) + (C)(0)|}{\sqrt{1+1+C^2} \cdot \sqrt{1+1}} \)
\( \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2} \cdot \sqrt{2+C^2}} \implies \sqrt{2+C^2} = 2 \implies 2+C^2 = 4 \implies C^2 = 2 \implies C = \pm \sqrt{2} \).
Equation of plane is \( -x - y \pm \sqrt{2}z + 1 = 0 \implies x + y \mp \sqrt{2}z - 1 = 0 \).

Question. Find the angle between the plane passing through point (1, 1, 1), (1, -1, 1), (-7, -3, -5) & x-z plane.
Answer: Let the plane be \( ax + by + cz + d = 0 \).
Using the points, the normal vector is found to be perpendicular to the y-axis.
The angle with the x-z plane (normal vector \( \hat{j} \)) is found to be \( \pi/2 \).

Question. Find the equation of the plane containing parallel lines \( (x - 4) = \frac{3 - y}{4} = \frac{z - 2}{5} \) and \( (x - 3) = \lambda (y + 2) = \mu z \).
Answer: The equation of plane containing parallel lines is of the form \( a(x - 4) + b(y - 3) + c(z - 2) = 0 \).
Using the given conditions for parallel lines, the coefficients are found:
\( \frac{a}{11} = \frac{b}{-1} = \frac{c}{-3} \).
Equation of plane: \( 11x - y - 3z = 35 \).

Question. Find the equation of image of the line \( \frac{x - 1}{9} = \frac{y - 2}{-1} = \frac{z + 3}{-3} \) in the plane \( 3x - 3y + 10z = 26 \).
Answer: The line and plane are parallel.
Image of a point (1, 2, -3) on the line about the plane \( 3x - 3y + 10z = 26 \) is \( (4, -1, 7) \).
The image line will be parallel to the original line.
Equation of image line: \( \frac{x - 4}{9} = \frac{y + 1}{-1} = \frac{z - 7}{-3} \).

Question. Find the distance between points of intersection of (i) Lines \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \) & \( \frac{x - 4}{5} = \frac{y - 1}{2} = z \) (ii) Lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) & \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \).
Answer: (i) Point of intersection \( P \) is found by equating general points: \( P(-1, -1, -1) \).
(ii) Point of intersection \( Q \) is found by equating the vector equations: \( Q(4, 0, -1) \).
Distance \( PQ = \sqrt{(4 - (-1))^2 + (0 - (-1))^2 + (-1 - (-1))^2} = \sqrt{5^2 + 1^2 + 0^2} = \sqrt{26} \).

Question. Find the equation of the sphere described on the line (2, -1, 4) and (-2, 2, -2) as diameter. Also find the area of the circle in which the sphere is intersected by the plane \( 2x + y - z = 3 \).
Answer: Diameter endpoints are (2, -1, 4) and (-2, 2, -2).
Center \( = (0, 1/2, 1) \).
Diameter \( = \sqrt{(2 - (-2))^2 + (-1 - 2)^2 + (4 - (-2))^2} = \sqrt{4^2 + (-3)^2 + 6^2} = \sqrt{16 + 9 + 36} = \sqrt{61} \).
Equation of sphere: \( (x - 0)^2 + (y - 1/2)^2 + (z - 1)^2 = \frac{61}{4} \)
\( \implies x^2 + y^2 + z^2 - y - 2z - 14 = 0 \).

Question. Find the plane \( \pi \) passing through the points of intersection of the planes \( 2x + 3y - z + 1 = 0 \) and \( x + y - 2z + 3 = 0 \) and is perpendicular to the plane \( 3x - y - 2z = 4 \). Find the image of point (1, 1, 1) in plane \( \pi \).
Answer: Let the required plane be \( \pi_1 + \lambda \pi_2 = 0 \).
\( (2x + 3y - z + 1) + \lambda(x + y - 2z + 3) = 0 \)
\( \implies (2 + \lambda)x + (3 + \lambda)y + (-1 - 2\lambda)z + (1 + 3\lambda) = 0 \).
Normal vector \( \vec{n} = (2+\lambda, 3+\lambda, -1-2\lambda) \).
It is perpendicular to \( 3x - y - 2z = 4 \) (normal \( \vec{n}_3 = (3, -1, -2) \)).
\( \vec{n} \cdot \vec{n}_3 = 0 \implies 3(2+\lambda) - (3+\lambda) - 2(-1-2\lambda) = 0 \)
\( \implies 6 + 3\lambda - 3 - \lambda + 2 + 4\lambda = 0 \implies 6\lambda + 5 = 0 \implies \lambda = -5/6 \).
Equation of plane \( \pi \): \( 7x + 13y + 4z - 9 = 0 \).

Question. Find the equation of the straight line which passes through the point (2, -1, -1); is parallel to the plane \( 4x + y + z + 2 = 0 \) and is perpendicular to the line of intersection of the planes \( 2x + y = 0 \), \( x - y + z \).
Answer: The equation of the line is \( \frac{x-2}{-1} = \frac{y+1}{13} = \frac{z+1}{9} \).

Question. If the distance between point \( (\alpha, 5\alpha, 10\alpha) \) from the point of intersection of the lines \( \vec{r} = (2\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 12\hat{k}) \) and plane \( \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \) is 13 units. Find the possible values of \( \alpha \).
Answer: Point of intersection of line and plane is \( (2, -1, 2) \).
Distance between \( (\alpha, 5\alpha, 10\alpha) \) and \( (2, -1, 2) \) is 13 units.
\( (\alpha - 2)^2 + (5\alpha + 1)^2 + (10\alpha - 2)^2 = 13^2 = 169 \).
Solving the quadratic equation in \( \alpha \):
\( \alpha = -1, \frac{80}{63} \).

Question. The edges of a rectangular parallelepiped are a, b, c; show that the angles between the four diagonals are given by \( \cos^{-1} \frac{\pm a^2 \pm b^2 \pm c^2}{a^2 + b^2 + c^2} \).
Answer: The direction ratios of the four diagonals are \( (a, b, c) \), \( (-a, b, c) \), \( (a, -b, c) \), and \( (a, b, -c) \).
The angle \( \theta \) between any two diagonals is given by \( \cos \theta = \frac{|\pm a^2 \pm b^2 \pm c^2|}{a^2 + b^2 + c^2} \).
Hence, \( \theta = \cos^{-1} \left( \frac{\pm a^2 \pm b^2 \pm c^2}{a^2 + b^2 + c^2} \right) \).

Question. Find the equation of the two lines through the origin which intersect the line \( \frac{x - 3}{2} = \frac{y - 3}{1} = \frac{z}{1} \) at an angle of \( \pi/3 \).
Answer: Let the line be \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \). Since it intersects the given line at origin, using the condition for coplanarity and the given angle:
The two lines are \( \frac{x}{1} = \frac{y}{2} = \frac{z}{-1} \) and \( \frac{x}{-1} = \frac{y}{1} = \frac{z}{-2} \).

Question. Find the equation of the projection of line \( 3x - y + 2z - 1 = 0, x + 2y - z - 2 = 0 \) on the plane \( 3x + 2y + z = 0 \).
Answer: The projection of the line on the plane is the line of intersection of the plane \( 3x + 2y + z = 0 \) and the plane containing the given line and perpendicular to it.
The required plane is found to be \( 3x - 8y + 7z + 4 = 0 \).
The equation of projection is: \( 3x - 8y + 7z + 4 = 0 \) and \( 3x + 2y + z = 0 \).

Question. Find the acute angle between the lines \( \frac{x - 1}{\ell} = \frac{y + 1}{m} = \frac{z}{n} \) & \( \frac{x + 1}{m} = \frac{y - 3}{n} = \frac{z - 1}{\ell} \) where \( \ell > m > n \) and \( \ell, m, n \) are the roots of the cubic equation \( x^3 + x^2 - 4x = 4 \).
Answer: \( x^3 + x^2 - 4x - 4 = 0 \implies (x+1)(x^2-4) = 0 \implies x = -1, 2, -2 \).
Since \( \ell > m > n \), we have \( \ell = 2, m = -1, n = -2 \).
Direction ratios of lines are \( (2, -1, -2) \) and \( (-1, -2, 2) \).
Angle \( \cos \theta = \frac{|(2)(-1) + (-1)(-2) + (-2)(2)|}{\sqrt{4+1+4} \cdot \sqrt{1+4+4}} = \frac{|-2 + 2 - 4|}{3 \cdot 3} = \frac{4}{9} \).
\( \implies \theta = \cos^{-1} \left( \frac{4}{9} \right) \).

Question. Let P(1, 3, 5) and Q(-2, 1, 4) be two points from which perpendiculars PM and QN are drawn to the x-z plane. Find the angle that the line MN makes with the plane \( x + y + z = 5 \).
Answer: Perpendiculars to x-z plane mean \( y \)-coordinate becomes 0.
\( M = (1, 0, 5) \) and \( N = (-2, 0, 4) \).
Direction ratios of MN are \( (-2-1, 0-0, 4-5) = (-3, 0, -1) \) or \( (3, 0, 1) \).
Plane normal is \( (1, 1, 1) \).
\( \sin \theta = \frac{|(3)(1) + (0)(1) + (1)(1)|}{\sqrt{3^2+0^2+1^2} \cdot \sqrt{1^2+1^2+1^2}} = \frac{4}{\sqrt{10} \cdot \sqrt{3}} = \frac{4}{\sqrt{30}} \).
\( \implies \theta = \sin^{-1} \left( \frac{4}{\sqrt{30}} \right) \).

Question. If 2d be the shortest distance between the lines \( \frac{y}{b} + \frac{z}{c} = 1 \); \( x = 0 \) and \( \frac{x}{a} - \frac{z}{c} = 1 \); \( y = 0 \) then prove that \( \frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \).
Answer: The lines are skew lines. Using the formula for shortest distance between skew lines, the distance \( 2d \) is calculated.
After derivation, we obtain the relation \( \frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \).

Question. Prove that the line \( \frac{x - 1}{2} = \frac{y - 2}{-3} = \frac{z + 3}{1} \) lies in the plane \( 3x + 4y + 6z + 7 = 0 \). If the plane is rotated about the line till the plane passes through the origin then find the equation of the plane in the new position.
Answer: Check if line lies in plane: \( 3(1) + 4(2) + 6(-3) + 7 = 3 + 8 - 18 + 7 = 0 \). It lies in the plane.
The new plane passes through the line and the origin.
The normal to the plane passing through origin and the line:
Direction of line is \( \vec{v} = (2, -3, 1) \). Point on line \( \vec{p} = (1, 2, -3) \).
Normal to new plane \( \vec{n} = \vec{v} \times \vec{p} = (2, -3, 1) \times (1, 2, -3) \).
\( \vec{n} = (9 - 2) \hat{i} - (-6 - 1) \hat{j} + (4 - (-3)) \hat{k} = 7\hat{i} + 7\hat{j} + 7\hat{k} \).
Equation of plane: \( 7x + 7y + 7z = 0 \implies x + y + z = 0 \)

Advanced Subjective Questions

Question. A line \( \frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z - k}{3} \) cuts the y-z plane and the x-y plane at A and B respectively. If \( \angle AOB = \frac{\pi}{2} \), then find k, where O is the origin.
Answer: Let a Point \( (\lambda - 2, 2\lambda + 3, 3\lambda + k) \)
In y-z plane x = 0
\( \implies \) \( \lambda = 2 \)
A (0, 7, 6 + k)
In x-y plane z = 0
\( \implies \) \( \lambda = -k/3 \)
\( B \left( -\frac{k}{3} - 2, -\frac{2k}{3} + 3, 0 \right) \)
\( \vec{OA} \cdot \vec{OB} = 0 \)
\( \implies \) \( 7 \left( \frac{-2k}{3} + 3 \right) = 0 \)
\( \implies \) \( k = \frac{9}{2} \)

Question. Find the volume of the tetrahedron with vertices P(2, 3, 2), Q(1, 1, 1), R(3, -2, 1) and S(7, 1, 4).
Answer: \( \vec{PQ} = (-1, -2, -1) \)
\( \vec{PR} = (1, -5, -1) \)
\( \vec{PS} = (5, -2, 2) \)
volume = \( \frac{1}{6} [\vec{PQ} \quad \vec{PR} \quad \vec{PS}] = \frac{1}{2} \)

Question. A sphere has an equation \( | \vec{r} - \vec{a} |^2 + | \vec{r} - \vec{b} |^2 = 72 \) where \( \vec{a} = \hat{i} + 3\hat{j} - 6\hat{k} \) and \( \vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k} \). Find
(i) the centre of the sphere
(ii) the radius of the sphere
(iii) perpendicular distance from the centre of the sphere to the plane \( \vec{r} \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = -3 \).
Answer: (i) \( (x - 1)^2 + (y - 3)^2 + (z + 6)^2 + (x - 2)^2 + (y - 4)^2 + (z - 2)^2 = 72 \)
\( \implies \) \( x^2 + y^2 + z^2 - 3x - 7y + 4z - 1 = 0 \)
Center \( \left( \frac{3}{2}, \frac{7}{2}, -2 \right) \)
(ii) \( r = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{7}{2}\right)^2 + (-2)^2 - (-1)} = \sqrt{\frac{33}{2}} \)
(iii) plane: \( 2x + 2y - z + 3 = 0 \)
\( d = \frac{|2(3/2) + 2(7/2) + 2 + 3|}{3} = 5 \)

Question. Let PM be the perpendicular from the point P(1, 2, 3) to the x-y plane. If OP makes an angle \( \theta \) with the positive direction of the z-axis and OM makes an angle \( \phi \) with the positive direction of the x-axis, where O is the origin, then find \( \theta \) and \( \phi \).
Answer: M (1, 2, 0)
\( \vec{OP} = (1, 2, 3) \)
\( \vec{OM} = (1, 2, 0) \)
\( \cos \theta = \frac{3}{\sqrt{14}} \)
\( \implies \) \( \theta = \cos^{-1} \frac{3}{\sqrt{14}} \)
\( \cos \phi = \frac{1}{\sqrt{5}} \)
\( \implies \) \( \phi = \cos^{-1} \frac{1}{\sqrt{5}} \)

Question. Prove that the line \( \frac{x}{1} = \frac{y}{1} = \frac{z - 1}{-2} \) lies in the plane x + y + z = 1. Find the lines in the plane through the point (0, 0, 1) which are inclined at an angle \( \cos^{-1} \left( \frac{1}{\sqrt{6}} \right) \) with the line.
Answer: \( \vec{a} = \text{DR's} = (1, 1, -2) \). Fixed point P(0, 0, 1).
x + y + z = 1 ; \( \vec{n} = (1, 1, 1) \)
\( \vec{n} \cdot \vec{a} = 1 + 1 - 2 = 0 \)
& point P satisfy the plane
\( \implies \) line lies is the plane.
Let the line \( \frac{x}{a} = \frac{y}{b} = \frac{z - 1}{c} \)
\( \cos \theta = \frac{1}{\sqrt{6}} = \frac{a + b - 2c}{\sqrt{a^2 + b^2 + c^2} \sqrt{6}} \)
squaring \( 3c^2 + 2ab = 4c(a + b) \) .........(1)
let the point is plane (1, 0, 0)
\( \implies \) condition of coplanarity \( \begin{vmatrix} 1 & 0 & -1 \\ a & b & c \\ 1 & 1 & -2 \end{vmatrix} = 0 \) ........(2)
Solve (1) & (2) and get (a, b, c)

Question. Find the equations of the straight line passing through the point (1, 2, 3) to intersect the straight line x + 1 = 2(y - 2) = z + 4 and parallel to the plane x + 5y + 4z = 0.
Answer: \( \frac{x - 1}{a} = \frac{y - 2}{b} = \frac{z - 3}{c} \)
\( \frac{x + 1}{2} = \frac{y - 2}{1} = \frac{z + 4}{2} \)
Lines are coplanar.
\( \begin{vmatrix} a & b & c \\ 2 & 1 & 2 \\ 2 & 0 & 7 \end{vmatrix} = 0 \)
\( \implies \) \( 7a - 10b - 2c = 0 \) ......(1)
and a + 5b + 4c = 0 ....(2)
from (1) & (2)
a = k, b = k, \( c = -\frac{3}{2}k \)
\( \frac{x - 1}{k} = \frac{y - 1}{k} = \frac{z - 3}{-\frac{3}{2}k} \)
\( \frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 3}{-3} \)

Question. Find the equations of the two lines through the origin which intersect the line \( \frac{x - 3}{2} = \frac{y - 3}{1} = \frac{z}{1} \) at an angle of \( \frac{\pi}{3} \).
Answer: \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \)
Lines will be coplanar so \( \begin{vmatrix} a & b & c \\ 2 & 1 & 1 \\ 3 & 3 & 0 \end{vmatrix} = 0 \)
\( \implies \) \( a = b + c \)
\( \cos 60^\circ = \frac{2a + b + c}{\sqrt{a^2 + b^2 + c^2} \sqrt{6}} \)
\( \implies \) \( 2b^2 + 2c^2 + 5bc = 0 \)
\( \implies \) \( (b + 2c) (2b + c) = 0 \)
b = -2c or b = -c/2
a = -c a = c/2
\( \frac{x}{-c} = \frac{y}{-2c} = \frac{z}{c} \) or \( \frac{x}{c/2} = \frac{y}{-c/2} = \frac{z}{c} \)
\( \frac{x}{1} = \frac{y}{2} = \frac{z}{-1} \) or \( \frac{x}{1} = \frac{y}{-1} = \frac{z}{2} \)

Question. Find the distance of the point P(-2, 3, -4) from the line \( \frac{x + 2}{3} = \frac{2y + 3}{4} = \frac{3z + 4}{5} \) measured parallel to the plane 4x + 12y - 3z + 1 = 0.
Answer: \( \frac{x + 2}{3} = \frac{y + 3/2}{2} = \frac{z + 4/3}{5/3} = \lambda \)
\( Q \left( 3\lambda - 2, 2\lambda - \frac{3}{2}, \frac{5\lambda - 4}{3} \right) \)
\( \vec{PQ} = \left( 3\lambda, 2\lambda - \frac{9}{2}, \frac{5\lambda + 8}{3} \right) \)
\( \vec{n} = (4, 12, -3) \)
\( \vec{PQ} \cdot \vec{n} = 0 \)
\( \implies \) \( \lambda = 2 \)
\( \vec{PQ} = \left( 6, -\frac{1}{2}, 6 \right) \)
distance = \( | \vec{PQ} | = \frac{17}{2} \)

Question. Find the equation to the line passing through the point (1, -2, -3) parallel to the line 2x + 3y - 3z + 2 = 0 = 3x - 4y + 2z - 4.
Answer: Direction of line = \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -3 \\ 3 & -4 & 2 \end{vmatrix} = -6\hat{i} - 13\hat{j} - 17\hat{k} \)
line: \( \frac{x - 1}{-6} = \frac{y + 2}{-13} = \frac{z + 3}{-17} \)
or \( \frac{x - 1}{6} = \frac{y + 2}{13} = \frac{z + 3}{17} \)

Question. Find the equation of the line passing through the point (4, -14, 4) and intersecting the line of intersection of the planes 3x + 2y - z = 5 and x - 2y - 2z = -1 at right angles.
Answer: \( \frac{x - 4}{a} = \frac{y + 14}{b} = \frac{z - 4}{c} \)
direction of intersecting line = \( \vec{n}_1 \times \vec{n}_2 = (-6, 5, -8) \)
Put z = 0 in both the planes
3x + 2y = 5
x - 2y = -1
x = 1, y = 1
P (1, 1, 0)
Another line \( \frac{x - 1}{-6} = \frac{y - 1}{5} = \frac{z - 0}{-8} \)
-6a + 5b - 8c = 0
both lines will be coplanar
\( \implies \) \( \begin{vmatrix} a & b & c \\ -6 & 5 & -8 \\ 3 & -15 & 4 \end{vmatrix} = 0 \)
\( \implies \) \( 4a = 3c \)
If a = k, \( c = \frac{4}{3}k, b = \frac{10}{3}k \)
\( \frac{x - 4}{3} = \frac{y + 14}{10} = \frac{z - 4}{4} \)

Question. Let P = (1, 0, -1); Q = (1, 1, 1) and R = (2, 1, 3) are three points.
(a) Find the area of the triangle having P, Q and R as its vertices.
(b) Given the equation of the plane through P, Q and R in the form ax + by + cz = 1.
(c) Where does the plane in part (b) intersect the y-axis.
(d) Give parametric equations for the line through R that is perpendicular to the plane in part (b).
Answer: (a) \( \vec{PQ} = (0, 1, 2) \), \( \vec{PR} = (1, 1, 4) \)
\( \vec{PQ} \times \vec{PR} = 2\hat{i} + 2\hat{j} - \hat{k} \)
Area = \( \frac{1}{2} | \vec{PQ} \times \vec{PR} | = \frac{3}{2} \)
(b) \( 2(x - 1) + 2(y - 0) - (z + 1) = 0 \)
2x + 2y - z - 3 = 0
\( \frac{2}{3}x + \frac{2}{3}y - \frac{1}{3}z = 1 \)
(c) x = 0, z = 0
\( \implies \) \( y = 3/2 \)
point \( \left( 0, \frac{3}{2}, 0 \right) \)
(d) dir of line will be along the normal of plane
\( \frac{x - 2}{2} = \frac{y - 1}{2} = \frac{z - 3}{-1} \)

Question. Find the point where the line of intersection of the planes x - 2y + z = 1 and x + 2y - 2z = 5, intersect the plane 2x + 2y + z + 6 = 0.
Answer: Direction of intersection line = \( \vec{n}_1 \times \vec{n}_2 \)
put z = 0 in both planes
x - 2y = 1, x = 3, y = 1
x + 2y = 5
point (3, 1, 0)
line : \( \frac{x - 3}{2} = \frac{y - 1}{3} = \frac{z - 0}{4} \)
variable point \( (2\lambda + 3, 3\lambda + 1, 4\lambda) \)
\( 2(2\lambda + 3) + 2(3\lambda + 1) + 4\lambda + 6 = 0 \)
\( \implies \) \( \lambda = -1 \)
point (1, -2, -4)

Question. Feet of the perpendicular drawn from the point P(2, 3, -5) on the axes of coordinates are A, B and C. Find the equation of the plane passing through their feet and the area of \( \Delta ABC \).
Answer: A(2, 0, 0) ; B(0, 3, 0) ; C(0, 0, -5)
normal of plane = \( \vec{AB} \times \vec{AC} \)
= (-15, -10, 6)
Equation of plane
-15(x - 2) - 10(y - 0) + 6(z - 0) = 0
\( \frac{x}{2} + \frac{y}{3} + \frac{z}{-5} = 1 \)
Area = \( \frac{1}{2} | \vec{AB} \times \vec{AC} | = \frac{19}{2} \)

Question. Find the equation to the line which can be drawn from the point (2, -1, 3) perpendicular to the lines \( \frac{x - 1}{2} = \frac{y - 2}{2} = \frac{z - 3}{2} \) and \( \frac{x - 4}{3} = \frac{y}{2} = \frac{z + 3}{1} \).
Answer: direction of line = \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 3 & 2 & 1 \end{vmatrix} = -2\hat{i} + 4\hat{j} - 2\hat{k} \)
equation of line
\( \frac{x - 2}{-2} = \frac{y + 1}{4} = \frac{z - 3}{-2} \)
\( \frac{x - 2}{1} = \frac{y + 1}{-2} = \frac{z - 3}{1} \)

Question. Find the equation of the plane containing the straight line \( \frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{z}{5} \) and perpendicular to the plane x - y + z + 2 = 0.
Answer: Normal of plane = \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 5 \\ 1 & -1 & 1 \end{vmatrix} = 2\hat{i} + 3\hat{j} + \hat{k} \)
equation of plane
2(x - 1) + 3(y - 2) + 1(z - 0) = 0
2x + 3y + z + 4 = 0

Question. Find the value of p so that the lines \( \frac{x - 1}{-3} = \frac{y - p}{2} = \frac{z + 2}{1} \) and \( \frac{x}{1} = \frac{y - 7}{-3} = \frac{z + 7}{2} \) are in the same plane. For this value of p, find the coordinates of their point of intersection and the equation of the plane containing them.
Answer: coplanar \( \implies \) \( \begin{vmatrix} -1 & p - 7 & -5 \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{vmatrix} = 0 \)
\( \implies \) \( p = 1 \)
\( \frac{x - 1}{-3} = \frac{y - 1}{2} = \frac{z + 2}{1} = \lambda \)
& \( \frac{x}{1} = \frac{y - 7}{-3} = \frac{z + 7}{2} = \mu \)
-3\(\lambda\) + 1 = \(\mu\) ....(1)
2\(\lambda\) + 1 = -3\(\mu\) + 7 ....(2)
\(\lambda\) - 2 = 2\(\mu\) - 7 ....(3)
\(\lambda\) = -3/7 & \(\mu\) = 16/7
Point of intersection \( \left( \frac{16}{7}, \frac{1}{7}, -\frac{17}{7} \right) \)
Normal of plane = \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{vmatrix} = (7, 7, 7) = (1, 1, 1) \)
Equation of plane (x - 1) + (y - 1) + (z + 2) = 0
x + y + z = 0

Question. Find the equations to the line of greatest slope through the point (7, 2, -1) in the plane x - 2y + 3z = 0 assuming that the axes are so placed that the plane 2x + 3y - 4z = 0 is horizontal.
Answer: \( \vec{n}_1 = (1, -2, 3) ; \vec{n}_2 = (2, 3, -4) \)
Direction of line = \( \vec{n}_1 \times (\vec{n}_1 \times \vec{n}_2) = (-44, -10, 8) \)
\( \frac{x - 7}{-44} = \frac{y - 2}{-10} = \frac{z + 1}{8} \)
or \( \frac{x - 7}{22} = \frac{y - 2}{5} = \frac{z + 1}{-4} \)

Question. Find the equation of the plane containing the line \( \frac{x - 1}{2} = \frac{y}{3} = \frac{z}{2} \) and parallel to the line \( \frac{x - 3}{2} = \frac{y}{5} = \frac{z - 2}{4} \). Find also the S.D. between the two lines.
Answer: Normal of plane = \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ 2 & 5 & 4 \end{vmatrix} = 2\hat{i} - 4\hat{j} + 4\hat{k} \)
plane will passes through (1, 0, 0)
\( \implies \) 1(x - 1) - 2y + 2z = 0
x - 2y + 2z = 1