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Point on a line at a given distance from a point on it (move along the unit direction)

This type of question, asked 4 times: 31 Jan 2024, 4 Apr 2024, 22 Jan 2025, 4 Apr 2025. All 4 questions below, oldest first.

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  1. Question 1

    31 Jan 2024

    178
    A line passes through A(4,−6,−2)A(4, - 6, - 2) and B(16,−2,4)B(16, - 2,4). The point P(a,b,c)P(a,b,c) where a,b,ca,b,c are non-negative integers, on the line ABAB lies at a distance of 21 units, from the point AA. The distance between the points P(a,b,c)P(a,b,c) and Q(4,−12,3)Q(4, - 12,3) is equal to

  2. Question 2

    4 Apr 2024

    76
    Let the point, on the line passing through the points P(1,−2,3)P(1, - 2,3) and Q(5,−4,7)Q(5, - 4,7), farther from the origin and at a distance of 9 units from the point PP, be (α,β,γ)(\alpha,\beta,\gamma). Then α2+β2+γ2\alpha^{2}+\beta^{2}+\gamma^{2} is equal to:

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  3. Question 3

    22 Jan 2025

    131
    Let a line pass through two distinct points P(−2,−1,3)P( - 2, - 1,3) and Q , and be parallel to the vector 3i^+2j^+2k3\widehat{i}+ 2\widehat{j}+ 2k. If the distance of the point QQ from the point R(1,3,3)R(1,3,3) is 5 , then the square of the area of △PQR\bigtriangleup PQR is equal to:

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  4. Question 4

    4 Apr 2025

    144
    Let A be the point of intersection of the lines L1:x−71=y−50=z−3−1L_{1}:\frac{x - 7}{1}=\frac{y - 5}{0}=\frac{z - 3}{- 1} and L2:x−13=L_{2}:\frac{x- 1}{3}= y+34=z+75\frac{y+ 3}{4}=\frac{z+ 7}{5}. Let BB and CC be the point on the lines L1L_{1} and L2L_{2} respectively such that AB=AC=15AB=AC=\sqrt{15}. Then the square of the area of the triangle ABC is :

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