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Distance of a point from a line measured along another line

This type of question, asked 9 times: 27 Jan 2024, 9 Apr 2024, 23 Jan 2025, 28 Jan 2025, 3 Apr 2025, 22 Jan 2026 S2, 28 Jan 2026 S1, 28 Jan 2026 S2, 4 Apr 2026 S1. All 9 questions below, oldest first.

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  1. Question 1

    27 Jan 2024

    62
    The distance, of the point (7,−2,11)(7, - 2,11) from the line x−61=y−40=z−83\frac{x - 6}{1}=\frac{y - 4}{0}=\frac{z - 8}{3} along the line x−52=y−1−3=z−56\frac{x - 5}{2}=\frac{y - 1}{- 3}=\frac{z - 5}{6}, is :

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  2. Question 2

    9 Apr 2024

    152
    Consider the line LL passing through the points (1,2,3)(1,2,3) and (2,3,5)(2,3,5). The distance of the point (113,113,193)\left( \frac{11}{3},\frac{11}{3},\frac{19}{3} \right) from the line LL along the line 3x−112=3y−111=3z−192\frac{3x - 11}{2}=\frac{3y - 11}{1}=\frac{3z - 19}{2} is equal to:

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  3. Question 3

    23 Jan 2025

    131
    The distance of the line x−22=y−63=z−34\frac{x - 2}{2}=\frac{y - 6}{3}=\frac{z - 3}{4} from the point (1,4,0)(1,4,0) along the line x1=y−22=z+33\frac{x}{1}=\frac{y - 2}{2}=\frac{z + 3}{3} is :

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  4. Question 4

    28 Jan 2025

    133
    The square of the distance of the point (157,327,7)\left( \frac{15}{7},\frac{32}{7},7 \right) from the line x+13=y+35=z+57\frac{x + 1}{3}=\frac{y + 3}{5}=\frac{z + 5}{7} in the direction of the vector i^+4j^+7k^\widehat{i}+ 4\widehat{j}+ 7\widehat{k} is :

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  5. Question 5

    3 Apr 2025

    144
    The distance of the point (7,10,11)(7,10,11) from the line x−41=y−40=z−23\frac{x- 4}{1}=\frac{y- 4}{0}=\frac{z- 2}{3} along the line x−92=y−133=z−176\frac{x- 9}{2}=\frac{y- 13}{3}=\frac{z- 17}{6} is.

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  6. Question 6

    22 Jan 2026 S2

    60
    Let L be the line x+12=y+13=z+36\frac{x + 1}{2}=\frac{y + 1}{3}=\frac{z + 3}{6} and let S be the set of all points (a,b,ca,b,c) on L , whose distance from the line x+12=y+13=z−90\frac{x + 1}{2}=\frac{y + 1}{3}=\frac{z - 9}{0} along the line L is 7. Then ∑(a,b,c)∈S(a+b+c)\sum_{(a,b,c) \in S}^{} (a + b + c) is equal to :

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  7. Question 7

    28 Jan 2026 S1

    68
    If the distances of the point (1,2,a)(1,2,a) from the line x−11=y2=z−11\frac{x- 1}{1}=\frac{y}{2}=\frac{z- 1}{1} along the lines L1:x−13=y−24=z−abL_{1}:\frac{x - 1}{3}=\frac{y - 2}{4}=\frac{z - a}{b} and L2:x−11=y−24=z−acL_{2}:\frac{x - 1}{1}=\frac{y - 2}{4}=\frac{z - a}{c} are equal, then a+b+ca + b + c is equal to

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  8. Question 8

    28 Jan 2026 S2

    74
    If the distance of the point P(43,α,β),β<0P(43,\alpha,\beta),\beta < 0, from the line r→=4i^−k^+μ(2i^+3k^),μ∈R\overrightarrow{r} = 4\widehat{i} - \widehat{k} + \mu\left( 2\widehat{i} + 3\widehat{k} \right),\mu \in R along aa line with direction ratios 3,−1,03, - 1,0 is 131013\sqrt{10}, then α2+β2\alpha^{2} + \beta^{2} is equal to ............ .

  9. Question 9

    4 Apr 2026 S1

    65
    The square of the distance of the point ( −2,−8,6- 2, - 8,6 ) from the line x−11=y−12=z−1\frac{x- 1}{1}=\frac{y- 1}{2}=\frac{z}{- 1} along the line x+51=y+5−1=z2\frac{x+ 5}{1}=\frac{y+ 5}{- 1}=\frac{z}{2} is equal to:

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