Vector Algebra and planes

Question

Given that $a = (3, 1, 2)$ and $b = (1, −2, −4)$ are the position vectors of the points $P$ and $Q$ respectively, find

(a) the equation of the plane passing through $Q$ and perpendicular to $PQ$ ;

(b) the distance from the point $(−1, 1, 1)$ to the plane obtained in (a).

I got the equation of the plane as $r•(2,3,6)=-28$ but I can't find the distance part b. The equation is correct and the distance is $5$ units but I don't know how they got that?

Thanks,

Answer 1

So you're looking for the distance of $(-1,1,1)$ to the plane with cartesian equation $$2x+3y+6z+28=0$$

1. Use a point-plane-distance formula: the distance from a point $(x_0,y_0,z_0)$ to the plane given by $ax+by+cz+d=0$ is equal to $$d = \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}= \ldots = 5$$

2. Do it "manually":

   • construct the line passing through $(-1,1,1)$ and perpendicular to the plane;
   • find the point of intersection of this line and the plane;
   • compute the distance between this point of intersection and the given point.

Answer 2

The distance from a point $(\alpha,\beta,\gamma)$ to a plane $\{(x,y,z)\,|\,ax+by+cz=d\}$ is$$\frac{\lvert a\alpha+b\beta+c\gamma-d\rvert}{\sqrt{a^2+b^2+c^2}}.$$