Question

Find an equation of the plane passing through the point (0,0,6) perpendicular to x=1-t y=2+t z=4-2t

Answer 1

Answer:

The equation of plane is x-y+z=12.

Step-by-step explanation:

It is given that the plane passing through the point (0,0,6) perpendicular to x=1-t y=2+t z=4-2t.

The given line is perpendicular to required plane, so coefficients of t represents the normal vector.

Normal vector is

$\overrightarrow {n}=$

If a plane passes through $(x_1,y_1,z_1)$ and having normal vector $\overrightarrow {n}=$, then the equation of plane is

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

$-1(x-0)+1(y-0)+(-2)(z-6)=0$

$-x+y-2z+12=0$

$-x+y-2z=-12$

Multiply both sides by -1.

$x-y+2z=12$

Therefore, the equation of plane is x-y+z=12.