- #1

RJLiberator

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## Homework Statement

Equation of the plane

Containing the line r(t) = <2 − t, 3, 4 + 2t> and point P(0, 0, 1).

## Homework Equations

ax+by+cz=d equation of a plane

## The Attempt at a Solution

1) We have a point, we need a normal vector

2) this time we are giving a line ON the plane, so the direction vector won't work as it is parallel to the plane.

3) Therefore, we can find the normal vector by using the point (0,0,1) and the point from the giving line (2, 3, 4) and making their line segment which comes out to be <-2, -3, -3>.

We then use this line segment and the direction vector <-1, 0, 2> take their cross product to net a vector that is perpendicular to the plane <6, -7, 3>.

And the equation of this plane becomes 6x-7y+3z=3

which satisfies both of our giving points (0, 0,1) and (2, 3, 4) on the plane.

The thing i am unsatisfied with here is why can we use 2 points on the plane, take their difference, and then cross it with the direction vector to get the perpendicular normal vector?

I understand that the cross product nets a perpendicular vector, but not the first part.