﻿ IIT - JEE Main Important Questions of Three Dimensional Geometry | Zigya

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# Three Dimensional Geometry

#### Multiple Choice Questions

1.

Let A (2, –3) and B(–2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line

• 2x + 3y = 9

• 2x – 3y = 7

• 3x + 2y = 5

• 3x – 2y = 3

A.

2x + 3y = 9

If C be (h, k) then centroid is (h/3, (k – 2)/3) it lies on 2x + 3y = 1.
⇒ locus is 2x + 3y = 9.

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2.

The curve y = (λ + 1)x2 + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals -

• {–2, 2}

• {1}

• {-2}

• {2}

C.

{-2}

for point of intersection at exactly one point
λx + 3 = (λ + 1)x2 + 2
(λ + 1)x2 – λx – 1 = 0
Δ = 0
λ2 + 4(λ + 1) = 0
λ2 + 4λ + 4 = 0
(λ + 2)2 = 0 λ = – 2

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3.

The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that

• it passes through the origin

• it makes angle π/2 + θ with the x-axis

• it passes through (aπ/2 ,-a)

• it is at a constant distance from the origin

D.

it is at a constant distance from the origin

Clearly dy/dx = an θ
⇒ slope of normal = - cot θ
Equation of normal at ‘θ’ is
y – a(sin θ - θ cos θ) = - cot θ(x – a(cos θ + θ sin θ)
⇒ y sin θ - a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.

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4.

If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the minimum value taken by c is

• 12

• 8

• 4

• 6

C.

4

The tangent of slope m must be of the form y = m(x + 2) + a/m

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5.

Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

• 3/2

• 5/2

• 7/2

• 9/2

C.

7/2

The distance between 4x + 2y + 4z - 16 = 0 and 4x + 2y + 4z + 5 = 0 is

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6.

If the straight lines x = 1 + s, y = –3 – λs, z = 1 + λs and x = t/ 2 , y = 1 + t, z = 2 – t with parameters s and t respectively, are co-planar then λ equals

• –2

• –1

• -1/2

• 0

A.

–2

Given   are coplanar then plan passing through these lines has normal perpendicular to these lines
⇒ a - bλ + cλ = 0 and a/2 +b -c =0 (where a, b, c are direction ratios of the normal to the plan) On solving, we get λ = -2.

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7.

A line makes the same angle θ, with each of the x and z-axis. If the angle β, which it makes with y-axis, is such that sin2β = 3sin2θ , then cos2θ equals

• 2/3

• 1/5

• 3/5

• 2/3

C.

3/5

A line makes angle θ with x-axis and z-axis and β with y-axis.
∴ l = cosθ, m = cosβ,n = cosθ
We know that, l2+ m2+ n2= 1

cos2θ + cos2β +cos2θ =1
2cos2θ = 1- cos2β
2cos2θ = sin2β
But sin2β = 3 sin2θ
therefore from equation (i) and (ii)
3sin2θ = 2cos2θ
3(1-cos2θ) = 2cos2θ
3-3cos2θ = 2cos2θ
3 = 5cos2θ

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8.

The intersection of the spheres x2 +y2 +z2 + 7x -2y-z =13 and x2 +y2 +z2 -3x +3y +4z = 8 is the same as the intersection of one of the sphere and the plane

• x-y-z =1

• x-2y-z =1

• x-y-2z=1

• 2x-y-z =1

D.

2x-y-z =1

Required plane is S1 – S2 = 0 where
S1 = x2 + y2 + z2 + 7x – 2y – z – 13 = 0
and S2 = x2 + y2 + z2 – 3x + 3y + 4z – 8 = 0
⇒ 2x – y – z = 1.

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9.

The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:

• x – 2y + 11 = 0

• x + 2y + 11 = 0

• x + 2y – 11 = 0

• x – 2y – 11 = 0

B.

x + 2y + 11 = 0

Equation of the required plane is
(x + y + z – 6) + λ(2x + 3y + z + 5) = 0
i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0
This plane is perpendicular to xy plane whose
equation is z = 0
i.e. 0 . x + 0 . y + z = 0
∴ By condition of perpendicularity
0 .(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0
i.e. λ = –1
∴ Equation of required plane is
(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) =0
or x + 2y + 11 = 0

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10.

A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by

• (3a, 3a, 3a), (a, a, a)

• (3a, 2a, 3a), (a, a, a)

• (3a, 2a, 3a), (a, a, 2a)

• (2a, 3a, 3a), (2a, a, a)

B.

(3a, 2a, 3a), (a, a, a)

Any point on the line  and any point on the line

Now direction cosine of the lines intersecting the above lines is proportional to (2t2 – a – t1, t2 – t1 + a, t2 – t1).
Hence 2t2 – a – t1 = 2k , t2 – t1 + a = k and t2 – t1 = 2k

On solving these, we get t1 = 3a , t2 = a. Hence points are (3a, 2a, 3a) and (a, a, a).

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