A plane is perpendicular to both [x,y,z] = [1, -10, 8] + s[1, 2, -1] and
[x,y,z] = [2, 5, -5] + t[2, 1, -3], and contains the point P(-1, 4, -2). Determine
if the point A(7, 10, 16) is also on this plane.
- I can't figure this one out. A vector perpendicular to the two lines found by cross product is [-5,1,0], but a plane requires two vectors. Can someone give me an insight as to how this problem can be solved? Thanks.
Best Answer
I only just now realized that you said perpendicular rather than parallel. A plane cannot be perpendicular to two lines going in different directions.
This problem only makes sense if the plane is parallel to the two given lines. Then the normal vector is
$[1,2,-1]\times[2,1,-3]=[-5,1,-3]$
and the equation of the plane is
$-5(x+1)+(y-4)-3(z+2)=0$
$5x-y+3z=-5$
$5(7)-10+3(16)\ne-5$
Therefore the point $A$ does not lie on the plane.