Find an equation for the plane that contains the line x=1+t,y=3t,z=2t and is parallel to the line of intersection of the planes −x+2y+z=0 and x+z+1=0
Here's the solution:
(−1,2,1)×(1,0,1)=(2,2,−2)
(2,2,-2)x(1,3,2)=(10,-6,4)
But my lecturer said the answer should be 5x-3y+2z-5=0 ?
On
Hint:
From the parametric equation of the line, you have instantly points on this line, and a directing vector $u$.
On another hand, from the equations of the planes, you have normal vectors of each of these planes, $n_1$ and $n_2$. A directing vector of their intersection line is $n_1\wedge n_2$.
Therefore, a normal vector for the plane you seek for, is $\;u\wedge(n_1\wedge n_2)=(u\cdot n_2)n_1-(u\cdot n_1)n_2$ by the double vector product formula.
This should be enough to obtain the equation of the plane.
Compute the cross product of $(-1,2,1)$ and $(1,0,1)$ which are the normal vectors to the intersecting planes. With the result, take cross product with $(1,3,2)$, the direction of the given line, to get the normal vector to your desired plane. Then find a point on the plane to get the full equation.