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 both lines to standard form.
In each fraction the variable must appear as $+x$, $+y$, $+z$ with coefficient $1$, then the denominators give the direction ratios.
Step 2: Fix the first line.
$\frac{1-x}{3}=\frac{x-1}{-3}$ and $\frac{7y-14}{2\lambda}=\frac{y-2}{\frac{2\lambda}{7}}$. So direction ratios of line 1 are $\left\langle -3,\ \frac{2\lambda}{7},\ 2\right\rangle$.
Step 3: Fix the second line.
$\frac{7-7x}{3\lambda}=\frac{x-1}{-\frac{3\lambda}{7}}$ and $\frac{6-z}{5}=\frac{z-6}{-5}$. So direction ratios of line 2 are $\left\langle -\frac{3\lambda}{7},\ 1,\ -5\right\rangle$.
Step 4: Use the right-angle condition.
Perpendicular lines have dot product zero: $(-3)\left(-\frac{3\lambda}{7}\right)+\left(\frac{2\lambda}{7}\right)(1)+(2)(-5)=0$.
Step 5: Simplify the equation.
$\frac{9\lambda}{7}+\frac{2\lambda}{7}-10=0$, so $\frac{11\lambda}{7}=10$.
Step 6: Solve for $\lambda$.
$11\lambda=70$, giving $\lambda=\frac{70}{11}$, which is option (2). \[ \boxed{\lambda=\frac{70}{11}} \]