Question

The equation of a line passing through $(3, -1, 2)$ and perpendicular to the lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda (2\hat{i} - 2\hat{j} + \hat{k})$ and $\bar{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \mu (\hat{i} - 2\hat{j} + 2\hat{k})$ is

• A. $\frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2}$
• B. $\frac{x - 3}{3} = \frac{y + 1}{2} = \frac{z - 2}{2}$
• C. $\frac{x + 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2}$
• D. $\frac{x - 3}{2} = \frac{y + 1}{2} = \frac{z - 2}{3}$

Hint:

Check the point coordinates first! A line passing through $(3, -1, 2)$ must have a numerator sequence of $(x-3)$, $(y+1)$, and $(z-2)$. This instantly eliminates option (C).

Answer 1

The Correct Option is A

Step 1: Understand the goal.
We want a line through $(3, -1, 2)$ that is perpendicular to both given lines.
Step 2: Read the two directions.
From the vector equations, the directions are $\bar{b}_1 = (2, -2, 1)$ and $\bar{b}_2 = (1, -2, 2)$.
Step 3: Get the new direction.
A line perpendicular to both has direction equal to the cross product $\bar{b}_1 \times \bar{b}_2$.
Step 4: Compute the cross product.
\[ \bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4+2) - \hat{j}(4-1) + \hat{k}(-4+2) \] \[ = -2\hat{i} - 3\hat{j} - 2\hat{k} \]
Step 5: Simplify the direction.
Multiply by $-1$: direction ratios $(2, 3, 2)$.
Step 6: Write the line.
Through $(3, -1, 2)$ with these ratios: \[ \frac{x-3}{2} = \frac{y+1}{3} = \frac{z-2}{2} \] \[ \boxed{\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-2}{2}} \]