Question

Find an equation for the plane which passes through the point P(1,3,1) and contains the line l: x=t, y=t, z=-2+t.

I know I need to find the normal vector and then plug in the values of the point to find D.

However, I'm not sure how to find the normal which is perpendicular to the given line.

Answer 1

youve already got one vector, the line direction

the line equation defines other points that you could use to define a vector to the stated point with

then cross the vectors to define the normal

Answer 2

the line vector is: 1,1,1

when t=0, a point on the line (in the same plane) is: 0,0,-2
the sated point is: 1,3,1

define a vector between the points to cross with

Answer 3

Oh! Thanks a ton!

From this I find line PQ=(-1,-3,-3)

Then I cross PQ x l = (0,-2,2) = N

This gives -2(y-3)+2(z-1)=0

Solve to find D.

-2y+2z+4=0

Factor out a 2, and divide by -1.

y-z-2=0

And that is the correct answer. Thanks a ton!