Q112 / 180 · KCET 2024 · Mathematics

If lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are mutually perpendicular, then $k$ is equal to

  • Correct answer
    A
    $-\frac{10}{7}$
  • B
    $-\frac{7}{10}$
  • C
    $-10$
  • D
    $-7$

Solution

$\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ $\therefore \quad \mathbf{b}_1=-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{k}$ and $\quad \mathbf{b}_2=3 k \hat{\mathbf{i}}+\hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ Now, $\quad \mathbf{b}_1 \cdot \mathbf{b}_2=\mathbf{0}$ $\Rightarrow \quad-9 k+2 k-10=0$ $\Rightarrow \quad-7 k=10$ $\therefore \quad k=\frac{-10}{7}$ Hence, the value of $k$ is $-\frac{10}{7}$

Hence, the correct answer is option: 'A'.

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