Question

If the lines $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to other, then $\lambda =$

• A. $\frac{-7}{6}$
• B. $\frac{6}{7}$
• C. $\frac{-6}{7}$
• D. $\frac{7}{6}$

Answer 1

Answer: (C)

$\frac{-6}{7}$

Answer 2

To find the value of $\lambda$, we first need to express the given lines in standard form and determine their direction vectors.

The first line is given as: $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$

We can rewrite this as: $\frac{2(x-2)}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$

Which simplifies to: $\frac{x-2}{\lambda/2}=\frac{y-1}{2}=\frac{z-3}{1}$

Thus, the direction vector for the first line, $\vec{d_1}$, is $\left(\frac{\lambda}{2}, 2, 1\right)$.

The second line is given as: $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$

We can rewrite this as: $\frac{x-1}{1}=\frac{3(y-\frac{1}{3})}{\lambda}=\frac{z-2}{1}$

Which simplifies to: $\frac{x-1}{1}=\frac{y-\frac{1}{3}}{\lambda/3}=\frac{z-2}{1}$

Thus, the direction vector for the second line, $\vec{d_2}$, is $\left(1, \frac{\lambda}{3}, 1\right)$.

Since the lines are perpendicular, their direction vectors are orthogonal, meaning their dot product is zero: $\vec{d_1} \cdot \vec{d_2} = 0$

So, we have: $\left(\frac{\lambda}{2}\right)(1) + (2)\left(\frac{\lambda}{3}\right) + (1)(1) = 0$

This simplifies to: $\frac{\lambda}{2} + \frac{2\lambda}{3} + 1 = 0$

Multiplying the entire equation by 6 to eliminate fractions: $3\lambda + 4\lambda + 6 = 0$

Combining terms: $7\lambda + 6 = 0$

Solving for $\lambda$: $7\lambda = -6$ $\lambda = \frac{-6}{7}$

Therefore, the value of $\lambda$ is $\frac{-6}{7}$.