Question

Obtain the equation of the plane passing through the line (x-1)/2 = -(y+1)/1 = (z-3)/4 and
which is perpendicular to the plane x+2y+z=4

Answer 1

Answer on Question #45088 – Analytic Geometry

Task:

Obtain the equation of the plane passing through the line $(x-1)/2 = -(y+1)/1 = (z-3)/4$ and which is perpendicular to the plane $x+2y+z=4$.

Solution:

As the plane is passing through the line $(x-1)/2 = -(y+1)/1 = (z-3)/4$, so the vector $(2,-1,4)$ is parallel to the plane. Since, our plane is perpendicular to the plane $x+2y+z=4$, then the vector $(1,2,1)$ is also parallel to our plane. It is easily seen that the point $(3,-2,7)$ belongs to line $(x-1)/2 = -(y+1)/1 = (z-3)/4$. Thus, our plane contains vectors $(x-3, y+2, z-7)$, $(1, 2, 1)$, $(2, -1, 4)$. The equation of plane can be found from the condition that three vectors are coplanar, which yields the determinant of the following matrix is zero.

$$= (x-3)(8+1) - (y+2)(4-2) + (z-7)(-1-4) = 9x - 2y - 5z + 4 = 0.$$

So, the equation of our plane is $9x - 2y - 5z + 4 = 0$

Answer: $9x - 2y - 5z + 4 = 0$

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