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Two lines are perpendicular: find the unknown

This type of question, asked 5 times: 12 May 2024 S1, 25 Apr 2025 S1, 15 Apr 2026 S1, 16 Apr 2026 S2, 15 May 2026 S1. All 5 questions below, oldest first.

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  1. Question 1

    12 May 2024 S1

    147
    If the lines 1−x3=7y−142p=z−32\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2} and 7−7x3p=y−51=6−z5\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5} are at right angles, then P=P=

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  2. Question 2

    25 Apr 2025 S1

    125
    If the lines 1−x2=7y+42λ=2z−52\frac{1 -x}{2}=\frac{7y+ 4}{2\lambda}=\frac{2z- 5}{2} and 7−7x3λ=y−17=6−z5\frac{7 - 7x}{3\lambda}=\frac{y- 1}{7}=\frac{6 -z}{5} are at right angle, then the value of λ\lambda is

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  3. Question 3

    15 Apr 2026 S1

    92
    If the lines x−55m+2=2−y5=1−z−1\dfrac{x - 5}{5m + 2} = \dfrac{2 - y}{5} = \dfrac{1 - z}{-1} and x=2y+14m=1−z−3x = \dfrac{2y + 1}{4m} = \dfrac{1 - z}{-3} are perpendicular to each other, then the value of m is ...

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  4. Question 4

    16 Apr 2026 S2

    93
    If the lines 2x=ky=−z2x = ky = -z and 6x=−y=−4z6x = -y = -4z are perpendicular to each other then the value of kk is ...

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  5. Question 5

    15 May 2026 S1

    93
    If the lines 1−x3=3y−12λ+1=4−z\dfrac{1 - x}{3} = \dfrac{3y - 1}{2\lambda + 1} = 4 - z and x+22μ=1−2y6=5−z7\dfrac{x + 2}{2\mu} = \dfrac{1 - 2y}{6} = \dfrac{5 - z}{7} are perpendicular, then 3μ+λ=…3\mu + \lambda = \ldots

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