Question

How do you find the vector perpendicular to the plane containing (-2,1,0), (-3,0,0), and (5,2,0)?

Answer 1

the Reqd. `bot` Vector`=(0,0,-6)=-6hatk`.

Explanation:

Let us name the given pts. contained in the plane, say `pi`,

`A(-3,0,0), B(-2,1,0) and C(5,2,0)`

Then, clearly, both `vec(AB) and, vec(AC) in pi`.

From Vector Geometry , we remember that the Cross Product

of two vectors is perpendicular to each of them.

Accordingly, `(vec(AB) xx vec(AC)) bot vec(AB) &, vec (AC)`, and,

hence, `(vec(AB) xx vec(AC)) bot pi`.

so, the Reqd. `bot` Vector`=(vec(AB) xx vec(AC))`

`=(1,1,0)xx(8,2,0)=(0,0,-6)=-6hatk`.

It is note-worthy that all the pts. `A, B, and, C` have `0` as their z-co-ordinate, so they all are in `XY`-plane, [i.e., plane `pi` is `XY`-plane], and, so, the reqd. vector `bot` to the plane `pi` has to be along `Z-axis`, i.e., along the vector`hatk`!