Question:
The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is
The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is
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To find a vector perpendicular to two others, use the Cross Product.
Updated On: Apr 30, 2026
- $\frac{x+1}{5} = \frac{y-2}{-4} = \frac{z+3}{1}$
- $\frac{x+1}{5} = \frac{y+2}{4} = \frac{z+3}{1}$
- $\frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{-1}$
- $\frac{x+1}{1} = \frac{y-2}{4} = \frac{z-3}{3}$
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The Correct Option is C
Solution and Explanation
Step 1: Find Direction Ratios
The line is perpendicular to both \( \vec{d_1} = (2, -3, -2) \) and \( \vec{d_2} = (-1, 2, 3) \).
\[ \vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -2 \\ -1 & 2 & 3 \end{vmatrix} \]
Step 2: Calculate Determinant
\[ \hat{i}(-9 + 4) - \hat{j}(6 - 2) + \hat{k}(4 - 3) \]
\[ (-5)\hat{i} - 4\hat{j} + \hat{k} = (-5, -4, 1) \]
Equivalently, \( (5, 4, -1) \).
Step 3: Form Equation of Line
Passing through \( (-1, 2, 3) \) with direction \( (5, 4, -1) \):
\[ \frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{-1} \]
Final Answer: (C)
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