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QUESTION: 1

If the distance of the point P(1, -2, 1) from the plane x + 2y - 2z = a, where a > 0 is 5, then foot of perpendicular from P to the plane is

Solution:

Distance of pt P from plane = 5

Gen. pt. on line is (λ + 1, 2λ - 2, -2λ + 1)

This pt line on plane

(λ + 1 + 2 2λ - 2) - 2 (-2λ + 1) = 10

QUESTION: 2

The image of point (3, -2, 1) in plane 3x – y + 4z = 2 is

Solution:

Let Q be the image of point P(3, -2, 1) PQ is normal to the plane

QUESTION: 3

The equation of plane passing through the point (0, 7, -7) and containing the line

Solution:

The equation of plane passing through (0, 7, -7) is a(x – 0) + b(y – 7) + c(z + 7) = 0

Plane contains line and passes through (-1, 3, -2)

∴ a(-1) + b(3 – 7) + c(-2 + 7) = 0

-3a + 2b + c = 0

∴ a : b : c = 1 : 1 : 1

x + y + z = 0

QUESTION: 4

Let the line lie in the plane x + 3y - az + b = 0, then (a, b) equals

Solution:

Line is perpendicular to the normal of plane

3 × 1 + (-5) × 3 + 2 (-α) = 0

– 12 – 2α = 0

α = – 6

Point (2, 1, -2) lie on plane

∴ b = 7

QUESTION: 5

The angle between line and plane 3x – 2y + 6z = 0 is (m is scalar)

Solution:

D. R’s of line are 2 : 1 : 2

D.R’s of Normal to plane is :3 : -2 : 6 Angle between line and plane

QUESTION: 6

Let P(3, 2, 6) be a point in space and Q be a point on line Then value of m for which the vector is parallel to the plane x-4y + 3z = 1 is

Solution:

QUESTION: 7

A variable plane at a distance of 1 unit from the origin cut the co-ordinate axis at A, B and C. If the centroid D (x, y, z) of ΔABC satisfy the relation then value of k is

Solution:

Equation of plane in intercept form is

Centroid of ΔABC(a/3, b/3, c/3)

It satisfies the relation

∴ 9/a^{2} + 9/b^{2} + 9/c^{2} = k

Perpendicular distance is 1 unit

QUESTION: 8

The Cartesian equation of plane passing through (1, 1, 1) and containing the x-axis is

Solution:

Eq. of x-axis is

Equation of plane passing through (1, 1, 1) and containing

QUESTION: 9

If the plane 2x – y + z = 0 is parallel to the line then value of a

is

Solution:

Plane is parallel to line i.e., normal of plane is perpendicular to the line

2 × 1 + (-1) × (-2) + 1 × a = 0

2 + 2 + a = 0

a = - 4

QUESTION: 10

Let L be the line of intersection of planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle ‘α’ with the positive x-axis, then cosa equals

Solution:

Let the direction cosines of line be

l, m, n then

2l + 3m + n = 0

And l + 3m + 2n = 0

On solving, l : m : n = 1 : - 1 : 1

l^{2} + m^{2} + n^{2} = 1

Line L make an angle ‘α’ with +ve x-axis

QUESTION: 11

The ratio in which the line joining (2, 4, 5), (3, 5, -4) is divided by the yz-plane is

Solution:

Let A(2, 4, 5) and B(3, 5, -4) and C be the point in which line AB meets the yz-plane. Let C divide AB in the ratio k : 1, then

But C lies on yz-plane, therefore,

QUESTION: 12

The equation of the plane through the point (-1, 2, 0) and parallel to the line

Solution:

P(-1,2,0) = -i + 2j

L1 = x/3 = (y+1)/0 = (z-2)/(-1)

a = 3i + 0j - k

L2 = (x-1)/1 = 2(y+½)/2 = 2(z+½)/(-1)

b = i + j - k/2

Plane is parallel to a ⇒ n is perpendicular to a

Plane is parallel to b ⇒ n is perpendicular to b

n = a * b = {(3, 0, -1) (1,1, -½)}

= i(0+1) -j(-3/2+1) +k(3-0)

= i + 1/2j + 3k

PR = R - P = (x+1)i + (y-z)j + zk

PR. n = 0

(x+1)*1 + (y-z)*½ + z*3 = 0

x+1 + y/2 - 1 + 3z = 0

x + y/2 + 3z = 0

2x + y + 6z = 0

QUESTION: 13

Equation of the line through (1, 1, 1) and perpendicular to the plane 2x + 3y – z – 5 = 0 is

Solution:

DR’s of the normal to the given plane are < 2, 3, -1 >.

Hence, the required line is

QUESTION: 14

Let A (1,1,1) , B (2, 3, 5) and C (-1, 0, 2) be three points, then equation of a plane parallel to the plane ABC which is a distance 2 is

Solution:

A (1,1,1), B (2,3,5), C(-1,0,2) direction ratios of AB are < 1,2,4>

Therefore, direction ratios of normal to plane ABC are <2, -3,1>

As a result, equation of the required plane is 2x – 3y +z = k then

Hence, equation of the required plane is

QUESTION: 15

The distance of the point (2, 1, -1) from the plane x - 2y + 4z = 9 is

Solution:

Required distance

QUESTION: 16

A variable plane is at a constant distance 3p form the origin and meets the axes in A, B and C. The locus of the centroid of the triangle ABC is

Solution:

Distance from origin = 3p

QUESTION: 17

The angle between the straight line and the plane 4x - 2y - 5z = 9

is

Solution:

**Correct Answer :- b**

**Explanation:- (x-1)/2=(y+3)/(-1)=(z-5)/2**

**4x - 2y - 5z = 9**

**cosθ=(2*4 - 1*2 + 2*(-5))/(3)√45)=0**

**= (8 - 2 -10)/3√45**

**= -4/(9√5)**

**Hence, angle between normal to plane and given line is 90°**

QUESTION: 18

The distance between the line and the plane

Solution:

Therefore, the line and the plane are parallel. A point on the line is (2, -2, 3). Required distance = distance of (2, -2, 3) from the given plane x + 5y + z - 5 = 0

QUESTION: 19

Equation of the plane containing the straight line and perpendicular to the

plane containing the straight line

Solution:

The Dr’s to normal of the plane containing

Let required plane is

From Eqs. (i) and (iv), required equation of plane, is x - 2 y + z =0

QUESTION: 20

If the distance between the plane x - 2y + z = d and the plane containing the line

Solution:

Equation of plane containing the given lines is

⇒ (x -1)(-1) - (y - 2)(-2) + (z - 3)(-1) = 0

⇒ -x +1+ 2y - 4 - z + 3 = 0

⇒ -x + 2 y - z = 0......(i)

Given plane is

x - 2 y + z = d .......(ii)

Eqs. (i) and (ii) are parallel

QUESTION: 21

If the lines intersect, then k is equal to

Solution:

Point of the first line (2t + 1, 3t - 1, 4t + 1)

Point of the second line (s + 3, 2s+ k, s) by comparing the points

QUESTION: 22

A line with positive direction cosines passes through the point P(2, -1, 2) and makes equal angle with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals

Solution:

∴ Any point on the line is

QUESTION: 23

The d.r’s of the line AB are 6, - 2,9. If the line AB makes angles α,β with oy, oz respectively where O = ( 0, 0, 0) then sin^{2}α - sin^{2}β =

Solution:

**Correct Answer :- A**

**Explanation:- Given, DR's of **

**AB =(6,−2,9)**

**cosα=(AB.i^)/|AB||i^|**

**= [6^2+2^2+9^2]^1/2**

** = 6/11**

**cosβ=−2/11**

**cosγ= 9/11**

**sin^2α−sin^2β=(1−cos^2α)−(1−cos^2β)**

**=cos^2β−cos^2α**

**=(4)/121 − (36)/121**

**= (−32)/121**

QUESTION: 24

A line makes the same angle θ with each of the x - axis and z - axis. It makes β angle with y - axis such that sin^{2} β = 3sin^{2} θ then cos^{2} θ

Solution:

QUESTION: 25

If the angle θ between the line and plane is such that Then, value of λ is

Solution:

QUESTION: 26

The line

Solution:

Direction cosines of the line are

⇒ (A) is correct

QUESTION: 27

A plane π passes through the point (1, 1, 1). If b, c, a are the direction ratios of a normal to the plane, where a, b, c ( a < b< c ) are the prime factors of 2001, then the equation of the plane π is

Solution:

QUESTION: 28

The point of intersection of the lines must be

Solution:

As the two lines intersect, we must have

a + a 'λ = a '+ μa,b + b 'λ = b '+μb and c + c 'λ = c'+ cμ

Solving a 'λ -μa = a'- a and b 'λ -μb = b'- b

We get, μ = 1 and λ = 1

⇒ x = a + a', y = b +b' and z = c+ c'

∴ Point of intersection is ( a + a ', b + b ', c + c ')

⇒ (B) is correct

QUESTION: 29

The direction ratios of a normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle of with the plane x + y = 3 are

Solution:

Any plane through (1,0,0) is

A (x -1) + B (y - 0) + C (z - 0) = 0 ..(1)

It contains (0,1,0) if -A + B = 0.... (2)

Also (1) makes an angle of with the plane x + y= 3

QUESTION: 30

The equation of the plane passing through the mid-point of the line joining the points (1, 2, 3) and (3, 4, 5) and perpendicular to it is:

Solution:

Direction ratio of line joining (1, 2, 3) and (3, 4, 5) are (2, 2, 2).

⇒ Normal to the plane has DR’s (2, 2, 2)

Now plane passes through

⇒ Equation of plane is

2(x – 2) + 2(y – 3) + 2(z – 4) = 0

x + y + z = 9

⇒ (A) is correct

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