Question

A plane P has a normal vector parallel to the direction vector of the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-1}{-1}$. The plane P passes through the point $(a, -2, 0)$ and the line passes through the point $(0, 1, 1)$. Find the value of $a$.

Answer 1

5

Answer 2

The direction vector of the line is $(1, 2, -1)$, which is the normal vector to the plane P. The equation of the plane is $x + 2y - z = d$. Since the line passes through $(0, 1, 1)$, we substitute this point into the plane equation: $0 + 2(1) - 1 = d \implies d = 1$. The equation of the plane is $x + 2y - z = 1$. Since the plane passes through $(a, -2, 0)$, we substitute these coordinates: $a + 2(-2) - 0 = 1 \implies a - 4 = 1 \implies a = 5$.