Question

Find the value of $\beta$ so that the line $\frac{x-2}{6}=\frac{y-1}{\beta }=\frac{z+5}{-4}$ is perpendicular to the plane $3x-y-2z=7$.

Answer 1

For a line and a plane,

angle

$sin\theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$

$(a_1,b_1,c_1)=(6,\beta ,-4)$

$(a_2,b_2,c_2)=(3,-1,-2)$

if they are perpendicular,

then, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\Rightarrow \frac{6}{3}=\frac{\beta }{-1}=\frac{-4}{-2}$

$\Rightarrow \beta =-2$