Question

If the two lines $\frac{x-1}{2} = \frac{1-y}{-a} = \frac{z}{4}$ and $\frac{x-3}{1} = \frac{2y-3}{4} = \frac{z-2}{2}$ are perpendicular, then the value of $a$ is equal to:

• A. $-4$
• B. $5$
• C. $-5$
• D. $4$
• E. $-2$

Hint:

Standardizing the equations is the most critical step. Be wary of terms like $1-y$ or $2y-3$; always ensure the coefficients of $x, y, z$ in the numerators are $+1$ before picking out the direction ratios.

Answer 1

The Correct Option is C

Concept: Two lines are perpendicular if the dot product of their direction vectors is zero. Before identifying the direction vectors, the line equations must be in the standard form: $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$.

Step 1: Rewrite the lines in standard form.
Line 1: $\frac{x-1}{2} = \frac{-(y-1)}{-a} = \frac{z}{4} \Rightarrow \frac{x-1}{2} = \frac{y-1}{a} = \frac{z}{4}$. Direction vector $\vec{b_1} = 2\hat{i} + a\hat{j} + 4\hat{k}$.
Line 2: $\frac{x-3}{1} = \frac{2(y-3/2)}{4} = \frac{z-2}{2} \Rightarrow \frac{x-3}{1} = \frac{y-3/2}{2} = \frac{z-2}{2}$. Direction vector $\vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k}$.

Step 2: Apply the perpendicular condition.
$\vec{b_1} \cdot \vec{b_2} = 0$: \[ (2)(1) + (a)(2) + (4)(2) = 0 \] \[ 2 + 2a + 8 = 0 \]

Step 3: Solve for $a$.
\[ 2a + 10 = 0 \] \[ 2a = -10 \quad \Rightarrow \quad a = -5 \]