Question

Find the equation of the plane which contains the line $x=2,y-z=0$ and is perpendicular to the plane $x+z=3.$ Find the point where this plane meets the line $\frac{x}{2}\equiv \frac{y}{3}=\frac{z}{1}$

Answer 1

$x=2$ ad $y-z=0$ line can be re-written as

$\frac{x-2}{0}=\frac{y}{1}=\frac{z}{1}$

Hence, its directional vector is

$j+k$...(i)

The required plane is perpendicular to the plane $x+z=0$.

Hence, the normal vector of the plane is perpendicular to the normal vector of the plane $x+z=0$.

Hence the required normal vector of the plane is

$\overrightarrow{n_1}\times \overrightarrow{n_2}$

$=(j+k)\times (i+k)$

$=i+j-k$

Hence, the required equation of the plane is

$x+y-z=d$

Now it contains the line $\frac{x-2}{0}=\frac{y}{1}=\frac{z}{1}$.

Hence, it must also contain the point $(2,0,0)$

Therefore the equation of the plane is

$x+y-z=2$

Consider the line

$\frac{x}{2}=\frac{y}{3}=\frac{z}{1}=t$

Hence, any point on the above line can be written as

$P=(2t,3t,t)$

Substituting in the equation of the plane, we get

$2t+3t-t=2$
$4t=2$
$t=0.5$

Hence $P=(1,1.5,0.5)$

$=(1,\frac{3}{2},\frac{1}{2})$