JEE Main Mathematics: Foot of the perpendicular and image of a point in a line Practice Questions
13 free multiple-choice questions on Foot of the perpendicular and image of a point in a line (Analytical Geometry) for JEE Main Mathematics, each with the correct answer and a step-by-step explanation. Attempt a question, then reveal the solution to check your reasoning.
Practice thousands more JEE Main questions in the app →Q1. The foot of the perpendicular from P(5, 1, −1) to the line (x − 1)/2 = y/1 = (z + 1)/2 is
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Correct answer: A — (3, 1, 1)
General point Q = (1 + 2t, t, −1 + 2t); PQ = (2t − 4, t − 1, 2t). PQ ⊥ line: 2(2t − 4) + (t − 1) + 2(2t) = 0, so 9t = 9 and t = 1. Foot = (3, 1, 1).
Q2. Assertion (A): The image of the point (1, 2, 3) in the line x = y = z is (3, 2, 1). Reason (R): The line x = y = z makes equal angles with the three coordinate axes.
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Correct answer: B — Both A and R are true but R is not the correct explanation of A
Foot from (1, 2, 3) to x = y = z: t = (1 + 2 + 3)/3 = 2, so F = (2, 2, 2). Image = 2F − P = (3, 2, 1), so A is true. R is true (DCs are all 1/√3), but it is not the reason for A; A follows from the foot of the perpendicular.
Q3. The foot of the perpendicular from P(1, 2, 6) to the line x = y = z is
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Correct answer: D — (3, 3, 3)
General point Q = (t, t, t); PQ = (t − 1, t − 2, t − 6). PQ ⊥ (1, 1, 1): 3t − 9 = 0, so t = 3. Foot = (3, 3, 3).
Q4. The image of the point (3, 2, 2) in the line x/1 = (y + 1)/1 = (z − 2)/(−2) is
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Correct answer: A — (−1, −2, −2)
General point Q = (t, −1 + t, 2 − 2t); PQ = (t − 3, t − 3, −2t). PQ ⊥ line: (t − 3) + (t − 3) + 4t = 6t − 6 = 0, so t = 1 and the foot is F(1, 0, 0). Image = 2F − P = (−1, −2, −2).
Q5. The perpendicular distance of the origin from the line (x − 3)/1 = (y + 1)/(−1) = (z − 2)/1 is
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Correct answer: C — √2
General point Q = (3 + t, −1 − t, 2 + t). OQ ⊥ line: (3 + t) + (1 + t) + (2 + t) = 0, so t = −2. Foot = (1, 1, 0); distance = √(1 + 1 + 0) = √2.
Q6. The foot of the perpendicular from A(8, 0, 3) to the line joining B(1, 1, 1) and C(3, −1, 2) is
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Correct answer: B — (5, −3, 3)
BC direction = (2, −2, 1); general point Q = (1 + 2t, 1 − 2t, 1 + t). AQ = (2t − 7, 1 − 2t, t − 2); AQ·(2, −2, 1) = 9t − 18 = 0, so t = 2. Foot = (5, −3, 3).
Q7. The foot of the perpendicular from the point (1, 2, 3) to the x-axis is
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Correct answer: C — (1, 0, 0)
Points on the x-axis are (x, 0, 0). The perpendicular from (1, 2, 3) meets it where x = 1, i.e. at (1, 0, 0).
Q8. The equation of the perpendicular drawn from P(3, 5, 0) to the line x/1 = (y − 1)/2 = (z + 1)/3 is
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Correct answer: B — (x − 3)/1 = (y − 5)/1 = z/(−1)
General point Q = (t, 1 + 2t, −1 + 3t); PQ = (t − 3, 2t − 4, 3t − 1). PQ ⊥ line: (t − 3) + 2(2t − 4) + 3(3t − 1) = 14t − 14 = 0, so t = 1 and Q = (1, 3, 2). PQ direction (−2, −2, 2) ∝ (1, 1, −1). Perpendicular: (x − 3)/1 = (y − 5)/1 = z/(−1).
Q9. The image of the point (3, −4, 5) in the x-axis is
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Correct answer: B — (3, 4, −5)
The foot on the x-axis is (3, 0, 0). Image = 2(3, 0, 0) − (3, −4, 5) = (3, 4, −5).
Q10. The image of the point P(3, −1, 3) in the line (x − 2)/1 = (y − 1)/2 = z/(−1) is
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Correct answer: D — (−1, −1, −1)
General point Q = (2 + t, 1 + 2t, −t); PQ = (t − 1, 2 + 2t, −t − 3). PQ ⊥ line: (t − 1) + 2(2 + 2t) + (t + 3) = 6t + 6 = 0, so t = −1 and the foot is F(1, −1, 1). Image = 2F − P = (2 − 3, −2 + 1, 2 − 3) = (−1, −1, −1).
Q11. P is the point (4, 3, 5) and Q is its image in the line r = (î + 2ĵ + 3k̂) + λ(2î − ĵ + 2k̂). The length PQ is
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Correct answer: C — 2√5
General point (1 + 2λ, 2 − λ, 3 + 2λ); P minus it = (3 − 2λ, 1 + λ, 2 − 2λ). Perpendicular: 2(3 − 2λ) − (1 + λ) + 2(2 − 2λ) = 9 − 9λ = 0, so λ = 1 and the foot is F(3, 1, 5). PF = √(1 + 4 + 0) = √5, so PQ = 2·PF = 2√5.
Q12. The image of the point P(3, 5, 1) in the line x = y = z is
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Correct answer: D — (3, 1, 5)
Foot: t = (3 + 5 + 1)/3 = 3, so F = (3, 3, 3). Image = 2F − P = (6 − 3, 6 − 5, 6 − 1) = (3, 1, 5).
Q13. The perpendicular distance of the point P(6, 3, 4) from the line (x − 1)/2 = (y − 2)/(−2) = (z − 3)/1 is
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Correct answer: B — 3√2
General point Q = (1 + 2t, 2 − 2t, 3 + t); PQ = (2t − 5, −1 − 2t, t − 1). PQ ⊥ line: 2(2t − 5) − 2(−1 − 2t) + (t − 1) = 9t − 9 = 0, so t = 1. Foot = (3, 0, 4); distance = √(9 + 9 + 0) = 3√2.
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