JEE Advanced (One or More Correct Option): Three dimensional Geometry | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. If a line makes angles α₁, α₂, α₃, α₄ with diagonals of a cube, then
(a)
(b)
(c)
(d)

Correct Answer is options (a, d)

Diagonals


Where, λ, m, n are direction cosine of line

Q.2. If a line makes an angle ‘Q’ with x and y axis then cot q can be equal to
(a) 1
(b) 2
(c) 0
(d) –3

Correct Answer is options (a, c)
cos²θ + cos²θ + cos²β = 1
⇒ cos²β = – cos2θ
⇒ cos2θ < 0, q∈ [π/4, π/2]
⇒ cot θ = [0, 1]

Q.3. Let PM be the perpendicular from the point P(1, 2, 3) to xy plane. If OP makes an angle θ with the +ve direction of the z-axis and OM makes an angle Ф with the positive direction of x-axis, where O is the origin then (θ and Ф are acute angles)
(a)
(b)
(c)
(d)

Correct Answer is options (a, b, c)
P be (x, y, z)
x = r sin θ. cos ϕ, y = r sin θ sin ϕ, z = r cos θ
1 = r sin θ. cos ϕ, 2 = r sin θ sinϕ, 3 = r cos θ
1² + 22 + 3² = r² sin² θ cos² ϕ + r² sin² θ sin² ϕ + r² cos² θ
= r² sin² θ (cos² ϕ + sin2 ϕ) + r² cos² θ
= r² sin² θ + r² cos² θ = r²
r² = 14
r = ± √14

Q.4. Let PN be perpendicular from point P(1, 2, 3) to xy plane if OP makes an angle α with positive direction of z- axis and ON makes an angle β with positive direction of x- axis, where O is origin (α, β are acute angles), then
(a)
(b)
(c)
(d) None of these

Correct Answer is options (a, c)
Let OP = r
∴1 = r sin α cosβ … (1)
2 = r sin α sin b … (2)
3 = r cos α … (3)
Square and add
r² = 14
r = ± √14 … (4)
Using (1), (2) (3) and (4), we get
(A) and (C) as answers.

Q.5. The equation of the line x + y + z – 1= 0, 4x + y – 2z + 2 = 0 written in the symmetrical form is
(a)
(b)
(c)
(d)

Correct Answer is options (a, b, c)
x + y + z – 1 = 0
4x + y – 2z + 2 = 0
∴ direction ratios of the line are (-3, 6, -3) i.e. (1, -2, 1)
Let  z = k, then x = k – 1, y = 2 – 2k
i.e. (k – 1, 2 – 2k, k) is any point on the line
∴ are points on the lines

Q.6. The co-ordinates of a point on the line  at a distance 4 √14 from the point (1, –1, 0) are -
(a) (9, –13, 4)
(b) (8 √14 + 1, –12 √14 – 1, 4√14 )
(c) (–7, 11, –4)
(d) (–8 √14 + 1, 12 √14 – 1, – 4 √14 )

Correct Answer is options (a, c)
The coordinates of a point on the given line are (2r + 1, – 3r – 1, r)
The distance of this point from the point (1, –1, 0) is given to be 4 √14 .
⇒
⇒ 14r² = 16 × 14 ⇒ r = ± 4
So the coordinates of the required point are
(9, – 13, 4) or (–7, 11, –4).

Q.7. If l₁ , m₁ , n₁ and l₂, m₂, n₂ are D.C.'s of the two lines inclined to each other at angle θ, then the D.C.'s of the internal and external bisectors of the angle between those lines are
(a)
(b)
(c)
(d) 

Correct Answer is options (b, c)
l₁l₂ + m₁m₂ + n₁n₂ = cosθ
Through origin O draw two lines parallel to given lines and take two points on each at a distance r from O and a point R on QO produced so that OR = r.
Then the co-ordinates of P, Q, R are (l₁r, m₁r, n₁r), (l₂r,m₂r, n₂r) and (-l₂r, -m₂r, -n₂r) respectively

If A, B be the mid points of PQ and PR then OA and OB are along the bisectors of the lines D.R.'s of
OA are l₁ + l₂, m₁ + m₂, n₁ + n₂ 
D.R.'s of OB are l₁ – l₂, m₁ – m₂, n₁ – n₂ 
Now ∑(l₁ + l₂)² = 1 + 1 + 2cosθ = 2(1 + cosθ) = 4cos² (θ/2)
and ∑(l₁ - l₂)² = 1 + 1 - 2cosθ = 2 (1 - cosθ) = 4sin²(θ/2)
∴ D.C. of internal and external bisectors are as given in (B) and (C) respectively.

Q.8. The equation of three planes are x-2y+ z = 3, 5x- y- z =8, and x+ y- z = 7 then
(a) they form a triangular prism
(b) all three plane have a common line of intersection
(c) line x/1 = y/2 = z/3 is parallel to each plane
(d) line x/1 = y/3 = z/4 intersect all three plane

Correct Answer is options (a, c)
Augment matrix
System of equation has no solution.
⇒ normals of plane are coplanar hence they are not intersecting at any point and forming a triangular prism.
(x, y, z ) = ( r, 2r, 3r ) does not satisfy by any plane for any value of 'r' hence x/1 = y/2 = z/3 is parallel to each plane. (x, y, z) = (r, 3r, 4r) satisfy by plane (1) & plane (2) for some value of ‘r’ but not satisfy by plane ‘3’ for any value of r. hence line x/1 = y/3 = z/4 does not interest plane ‘3’.

Q.9. The plane passing through the origin and containing the lines whose direction cosines are proportional to 1, –2, 2 and 2, 3, –1 passes through the point, is  
(a)  (1, –2, 2)
(b) (2, 3, –1)
(c) (3, 1, 1)
(d) (4, 0, 7) 

Correct Answer is options (a, b, c)
Let the direction cosines of the normal to the plane be l, m, n . Since the normal is perpendicular to both the lines lying in the plane, whose direction cosines are proportional to (1, –2, 2) and (2, 3, –1), we have
l – 2m + 2n  = 0  &  2l + 3m – n = 0
⇒
Since the plane passes through the origin its equation is
lx + my + nz = 0 or 4x – 5y – 7z = 0.

Q.10. If a line makes an angle ‘Q’ with x and y axis then cot q can be equal to
(a) 1
(b) 2
(c) 0
(d) –3

Correct Answer is options (a, c)
cos²θ + cos²θ + cos²β = 1
⇒ cos²β = – cos2θ
⇒ cos2θ < 0, q∈ [π/4, π/2]
⇒ cot θ = [0, 1]