Finding a plane perpendicular to two lines and a point

Question

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.

Answer 1

Since the plane is perpendicular to the given lines, then the direction vector of these lines are normal to the plane. Take, for instance, the vector $\vec{N} = (1,2,-1)$. We know the point $P(-1,4,-2)$ lies in the plane. Pick arbitrary $X(x,y,z)$ in our plane. Then, the equation of our plane is given by

$$ \vec{N} \cdot \vec{PX} = 0 \iff (1,2,-1)\cdot(x+1,y-4,z+2)=0$$

Which is equivalent to $x+2y-z-11=0$. Now, $7+2(10)-16-11 = 0$, which means that the point $(7,10,16)$ lies in the plane.

Answer 2

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.