The equation of planes bisecting the angle between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is / are

Q: The equation of planes bisecting the angle between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is / are

(A) 5x – y – 4z – 45 = 0

(B) 5x – y – 4z – 3 = 0

(C) 23x – 13y + 32z + 45 = 0

(D) 23x – 13y + 32z + 5 = 0

Sol. Equation of planes bisecting the angle between given planes are

$\large \frac{2x-y+ 2z+3}{\sqrt{4+1+4}} = \pm \frac{3x-2y+6z+8}{\sqrt{9+4+36}}$

⇒ 5x – y – 4z – 3 = 0 and 23x – 13y + 32z + 45 = 0 are required planes.

Hence (B) and (C) are the correct answers.

The equations of the lines of shortest distance between the lines …

Q: The equations of the lines of shortest distance between the lines $\large \frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$ and $\large \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ are

(A) 3(x – 21) = 3y + 92 = 3z – 32

(B) $\large \frac{x-62/3}{1/3} = \frac{y+31}{1/3} = \frac{z-31}{1/3}$

(C) $\large \frac{x-21}{1/3} = \frac{y + 92/3}{1/3} = \frac{z-32/3}{1/3}$

(D) $\large \frac{x-2}{1/3} = \frac{y+3}{1/3} = \frac{z-1}{1/3}$

Sol. Let P(2r1 , – 3r1, r1) and Q(3r2 + 2, – 5r2 + 1, 2r2 – 2) be the points on the given lines so that PQ is the line of shortest distance

d.r.s of PQ 2r1 – 3r2 – 2 , – 3r1 + 5r2 – 1, r1 – 2r2 + 2

Since it is perpendicular to given lines

2(2r1 – 3r2 – 2) – 3(3r1 + 5r2 – 1) + (r1 – 2r2 + 2) = 0

and (2r1 – 3r2 – 2) – 5(– 3r1 + 5r2 – 1) + 2(r1 – 2r2 + 2) = 0

r1 = 31/3, r2 = 19/3

P is (62/3 , -31 , 31/3) and Q is (21 , -92/3 , 32/3)

d.r.s of PQ is $(\frac{1}{3} , \frac{1}{3} , \frac{1}{3})$

Hence (A), (B) and (C) are the correct answers.

The lines $\frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3}$

Q: The lines  $\large \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3}$

(A) are non coplanar

(B) are coplanar

(C) intersecting at (4, 0, – 1)

(D) intersecting at (1, 1, – 1)

Sol: Let $\large \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} = \lambda_1$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3} = \lambda_2$

Then 1 + 3λ1 = 4 + 2λ2 …(1)

1 – λ1 = 0 …(2)

– 1 = 3λ2 – 1 …(3)

λ1 = 1 , λ2 = 0 satisfies (1)

lines are intersecting hence coplanar and point of intersection (4, 0, – 1).

Hence (B) and (C) are the correct answers.

The coordinate of the points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ which are at a distance 3√2 from the point (1, 2, 3)

Q: The coordinate of the points on the line $\large \frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ which are at a distance 3√2 from the point (1, 2, 3)

(A) (– 2, – 1, 3)

(B) (2, 2, 4)

(C) $( \frac{56}{17} , \frac{43}{17} , \frac{111}{17} )$

(D) $( \frac{47}{11} , \frac{42}{11} , \frac{56}{11} )$

Sol. Any point on the line (3λ – 2, 2λ – 1, 2λ + 3), then

(3λ – 3)2 + (2λ – 3)2 + (2λ)2 = 18

λ = 0, 13/17

points are (– 2, – 1, 3) and $( \frac{56}{17} , \frac{43}{17} , \frac{111}{17} )$

Hence (A) and (C) are the correct answers.

The equation of the straight line through the origin parallel to the line (b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

Q: The equation of the straight line through the origin parallel to the line
(b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

(A) $\large \frac{x}{b^2-c^2} = \frac{y}{c^2-a^2} = \frac{z}{a^2-b^2}$

(B) $\large \frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

(C) $\large \frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab}$

(D) none of these

Sol. Equation of straight line through origin is

$\large \frac{x-0}{l} = \frac{y-0}{m} = \frac{z-0}{n}$

where l((b+c) + m(c+a) + n(a + b) = 0

and l(b – c) + m(c -a) + n(a – b) = 0

On solving,

$\large \frac{l}{2(a^2-bc)} = \frac{m}{2(b^2-ca)} = \frac{n}{2(c^2-ab)}$

Equation of the straight line is

$\large \frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab}$

Hence (C) is the correct answer.

A straight line passes through the point (2, –1, –1). It is parallel to the plane  4x + y + z + 2 = 0 and …

Q: A straight line passes through the point (2, –1, –1). It is parallel to the plane  4x + y + z + 2 = 0 and is perpendicular to the line $\frac{x}{1} = \frac{y}{-2} = \frac{z-5}{1}$ . The equation of the straight line is

(A) $\large \frac{x-2}{4} = \frac{y+1}{1} = \frac{z+1}{1}$

(B) $\large \frac{x+2}{4} = \frac{y-1}{1} = \frac{z-1}{1}$

(C) $\large \frac{x-2}{-1} = \frac{y+1}{1} = \frac{z+1}{3}$

(D) $\large \frac{x+2}{-1} = \frac{y-1}{1} = \frac{z-1}{3}$

Sol. Let direction cosines of straight line be l, m, n

∴ 4l + m + n = 0

l – 2m + n = 0

$\frac{l}{3} = \frac{m}{-3} = \frac{n}{-9}$

$\frac{l}{-1} = \frac{m}{1} = \frac{n}{3}$

Equation of straight line is

$\large \frac{x-2}{-1} = \frac{y+1}{1} = \frac{z+1}{3}$

Hence (C) is the correct answer.

The shortest distance between the two straight lines…

Q: The shortest distance between the two straight lines $\frac{x-4/3}{2} = \frac{y + 6/5}{3} = \frac{z-3/2}{4}$ and $\frac{5y+6}{8} = \frac{2z-3}{9} = \frac{3x-4}{5}$ is

(A) $\sqrt{29}$

(B) 3

(C) 0

(D) $6 \sqrt{10}$

Sol. Since these two lines are intersecting so shortest distance between the lines will be 0.

Hence (C) is the correct answer.

The equation of the plane bisecting the acute angle between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is

Q: The equation of the plane bisecting the acute angle between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is

(A) 23x – 13y + 32z + 45 = 0

(B) 5x – y – 4z = 3

(C) 5x – y – 4z + 45 = 0

(D) 23x – 13y + 32z + 3 = 0

Sol. a1a2 + b1 b2 + c1c2 = 6 + 2 + 12 =positive

So acute angle bisector is $\large \frac{2x – y + 2z +3}{3} = – \frac{3x-2y + 6z + 8 }{7}$

⇒ 23x – 13y + 32z + 45 = 0

Hence (A) is the correct answer.

Consider the following statements: Assertion (A): the plane y + z + 1 = 0 is parallel to x-axis. Reason (R): normal to the plane is parallel to x-axis.

Q: Consider the following statements:
Assertion (A): the plane y + z + 1 = 0 is parallel to x-axis.
Reason (R): normal to the plane is parallel to x-axis.
Of these statements:

(A) both A and R are true and R is the correct explanation of A

(B) both A and R are true and R is not a correct explanation of A

(C) A is true but R is false

(D) A is false but R is true

Sol: Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0

but normal to the plane will be perpendicular to x-axis.

Hence (C) is the correct answer.

A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C…

Q: A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the centroid D (x, y, z) of triangle ABC satisfies the relation $\large \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$ , then the value of k is

(A) 3

(B) 1

(C) 1/3

(D) 9

Sol: Let $\large \frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1$ be the variable plane so that

$\large \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1$

The plane meets the coordinate axes at A (a, 0, 0), B (0, b, 0), C (0, 0, c). The centroid D of the triangle ABC is $(\frac{a}{3} , \frac{b}{3} , \frac{c}{3})$

Þ x = a/3 , y = b/3 , z = c/3

and $\large \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1$

$\large \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9$

Hence (D) is the correct answer