Question

If the lines $\frac{1-x}{3}=\frac{7y-14}{2\lambda}=\frac{z-3}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles, then $\lambda =$

• A. $-\frac{70}{11}$
• B. $\frac{70}{11}$
• C. $\frac{11}{70}$
• D. $-\frac{11}{70}$

Hint:

The most common mistake in 3D geometry problems is immediately extracting direction ratios from non-standard line equations! Always guarantee the coefficients of $x$, $y$, and $z$ in the numerator are exactly $+1$ before pulling your denominator values.

Answer 1

The Correct Option is B

Step 1: Bring each line to standard form.
For line 1 the direction ratios are $\langle -3, \tfrac{2\lambda}{7}, 2\rangle$. For line 2 they are $\langle \tfrac{-3\lambda}{7}, 1, -5\rangle$.

Step 2: Use the right angle condition.
Perpendicular lines have a zero dot product:
$$(-3)\!\left(\tfrac{-3\lambda}{7}\right) + \tfrac{2\lambda}{7}(1) + 2(-5) = 0$$

Step 3: Solve for $\lambda$.
$$\frac{9\lambda}{7} + \frac{2\lambda}{7} - 10 = 0 \Rightarrow \frac{11\lambda}{7} = 10 \Rightarrow \lambda = \frac{70}{11}$$
\[ \boxed{\dfrac{70}{11}} \]