Three-Dimensional Analytical Geometry: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Lines in symmetric form ax−x1=by−y1=cz−z1 have direction ratios (a,b,c); a zero denominator means the corresponding numerator is 0. For real k, let L1:2x−1=ky+1=4z−2 and L2:1x−3=2y−1=2z−k. The set of all real k for which L1 and L2 intersect in exactly one point is ______. Exactly one option is correct.
A{4,6}
B{6}
C{4}
D{−6,4}
Show answer and solution
Answer: B
L1 passes through A=(1,−1,2) with direction d1=(2,k,4); L2 passes through B=(3,1,k) with direction d2=(1,2,2).
Coplanarity: 2212k2k−242=0 (rows AB, d1, d2). Expanding gives −(k−4)(k−6)=0, so the lines are coplanar exactly for k=4 or k=6.
k=4: d1=(2,4,4)=2d2, so the lines are parallel. B=(3,1,4) is not on L1 (it would need 23−1=41+1), so they are distinct parallel lines: coplanar but with no common point.
k=6: the directions are not parallel. Solving (1+2s,−1+6s,2+4s)=(3+t,1+2t,6+2t) gives s=−1, t=−4, common point (−1,−7,−2).
Hence the lines meet in exactly one point only for k=6: the set is {6}.
Trap: the determinant condition says only "coplanar"; the parallel root k=4 must be thrown out.
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
A line L through the origin makes an angle 3π with the positive x-axis and an angle 4π with the positive y-axis. The angle between a line and a plane is taken in [0,2π]. Over all such lines L, the set of all possible values of the angle between L and the plane x−z=5 is ______. Exactly one option is correct.
A{4π}
B{0}
C{6π,4π}
D{0,4π}
Show answer and solution
Answer: D
Direction cosines: l=cos3π=21, m=cos4π=21, and n2=1−41−21=41, so n=±21. Two lines are possible.
The plane has normal (1,0,−1). If φ is the angle between L and the plane, sinφ=2∣l−n∣.
n=21: sinφ=0, so φ=0. The line is parallel to the plane; it does not lie in it, since the origin is not on x−z=5.
n=−21: sinφ=21, so φ=4π.
Set of values: {0,4π}.
Difficulty: Hard
Question 3Level 3 · AdvancedSingle correct
Lines in symmetric form ax−x1=by−y1=cz−z1 have direction ratios (a,b,c); a zero denominator means the corresponding numerator is 0. Let L:−1x=1y−2=1z+2 and let Π be the plane x+2y+2z=9. The image (mirror reflection) of L in Π is a line L′. The point where L′ meets the plane z=0 is ______. Exactly one option is correct.
A(−8,4,0)
B(1,4,0)
C(−2,4,0)
D(7,8,0)
Show answer and solution
Answer: A
L passes through A=(0,2,−2) with direction (−1,1,1). The direction is not perpendicular to the normal (1,2,2) (−1+2+2=3=0), so L meets Π.
Meeting point: (−t,2+t,−2+t) satisfies −t+4+2t−4+2t=9, so t=3 and Q=(−3,5,1). Q is fixed by the reflection, so it lies on L′.
Image of A: 91⋅0+2⋅2+2⋅(−2)−9=−1, so A′=A−2(−1)(1,2,2)=(2,6,2).
L′ passes through Q(−3,5,1) and A′(2,6,2), with direction (5,1,1): (x,y,z)=(−3+5u,5+u,1+u).
z=0 gives u=−1, so the point is (−8,4,0).
Trap: (1,4,0) is the foot of the perpendicular from A. It is where the projection of L meets z=0, not the image. (−2,4,0) is where L itself meets z=0.
Difficulty: Very hard
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