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Unit vector or direction cosines perpendicular to two given lines

This type of question, asked 6 times: 3 May 2023 S2, 4 May 2023 S1, 15 May 2023 S1, 14 May 2024 S1, 15 Apr 2026 S1, 15 May 2026 S1. All 6 questions below, oldest first.

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  1. Question 1

    3 May 2023 S2

    117
    Let L1:x+13=y+22=z+11L_1:\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1} and L2:x−22=y+21=z−33L_2:\frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3} be the given lines. Then the unit vector perpendicular to L1L_1 and L2L_2 is

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  2. Question 2

    4 May 2023 S1

    129
    Let L1:x+13=y+21=z+12L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2} and L2:x−21=y+22=z−33L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3} be two given lines. Then the unit vector perpendicular to L1L_1 and L2L_2 is

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  3. Question 3

    15 May 2023 S1

    128
    Let L1:x+25=y−32=z−61L_{1}:\frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1} and L2:x−34=y+23=z−35L_{2}:\frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5} be the given lines. Then the unit vector perpendicular to both L1L_{1} and L2L_{2} is

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  4. Question 4

    14 May 2024 S1

    116
    Consider the lines L1:x+13=y+21=z+12L_1:\dfrac{x+1}{3}=\dfrac{y+2}{1}=\dfrac{z+1}{2} and L2:x−21=y+22=z−33L_2:\dfrac{x-2}{1}=\dfrac{y+2}{2}=\dfrac{z-3}{3}, then the unit vector perpendicular to both L1L_1 and L2L_2 is

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  5. Question 5

    15 Apr 2026 S1

    94
    The direction cosines of a line which is perpendicular to the lines x−72=y+17−3=z−61\dfrac{x - 7}{2} = \dfrac{y + 17}{-3} = \dfrac{z - 6}{1} and x+51=y+32=z−6−2\dfrac{x + 5}{1} = \dfrac{y + 3}{2} = \dfrac{z - 6}{-2} are...

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  6. Question 6

    15 May 2026 S1

    94
    Consider the lines L1:x2=y2=−zL_1: \dfrac{x}{2} = \dfrac{y}{2} = -z, L2:x−11=y−12=z−32L_2: \dfrac{x - 1}{1} = \dfrac{y - 1}{2} = \dfrac{z - 3}{2}. The unit vector perpendicular to both L1L_1 and L2L_2 is:

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