Question
Find the value of $\lambda$ such that the line $\frac{\text{x}-2}{6}=\frac{\text{y}-1}{\lambda}=\frac{\text{z}+5}{-4}$ is perpendicular to the plane 3x - y - 2z = 7.
✓
Answer
Here, given mid line $\frac{\text{x}-2}{6}=\frac{\text{y}-1}{\lambda}=\frac{\text{z}+5}{-4}$ is parpendicular to plane 3x - y - 2z = 7 so, normal vector of plane is parallel to line so,
Direction ratios of normal to plane are proparional to the direction ratios of line.
Here,
$\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}$
cross multiplying the last two
$-2\lambda=4$
$\lambda=\frac{4}{-2}$
$\lambda=-2$
Direction ratios of normal to plane are proparional to the direction ratios of line.
Here,
$\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}$
cross multiplying the last two
$-2\lambda=4$
$\lambda=\frac{4}{-2}$
$\lambda=-2$
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