Question

The projection of the line $\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$ on a plane P is $\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1}$. If plane P passes through $(a, -2, 0)$, then $a$ is equal to

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (C)

5

Answer 2

The direction vector of the first line is $\vec{d_1} = \langle 2, 2, 1 \rangle$. The direction vector of the projected line is $\vec{d_2} = \langle 1, 1, -1 \rangle$. The normal vector to the plane P, $\vec{n}$, is parallel to $\vec{d_1} - \vec{d_2} = \langle 1, 1, 2 \rangle$. Thus, the equation of the plane is $x + y + 2z + D = 0$. A point on the projected line is $(0, 1, 1)$. Substituting this into the plane equation gives $0 + 1 + 2(1) + D = 0$, so $D = -3$. The plane equation is $x + y + 2z - 3 = 0$. Since the plane passes through $(a, -2, 0)$, we have $a + (-2) + 2(0) - 3 = 0$, which gives $a = 5$.