Question

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point (0, 7, -7) is

• A. 2x + y + z = 0
• B. x + y + z = 0
• C. x + 2y - 3z = 35
• D. x + 3y + z = 14

Answer 1

Answer: (B)

x + y + z = 0

Answer 2

To find the equation of the plane, we can follow these steps:

1. Find a point on the line and its direction vector:

   The given line is $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$. A point on this line is $Q(-1, 3, -2)$, and the direction vector is $\vec{d} = (-3, 2, 1)$.

2. Find another vector in the plane:

   We are given the point $P(0, 7, -7)$. The vector from $Q$ to $P$ is $\vec{QP} = P - Q = (0 - (-1), 7 - 3, -7 - (-2)) = (1, 4, -5)$.

3. Find the normal vector to the plane:

   The normal vector $\vec{n}$ is the cross product of $\vec{d}$ and $\vec{QP}$: $\vec{n} = \vec{d} \times \vec{QP} = (-3, 2, 1) \times (1, 4, -5)$.

   Calculating the cross product: $\vec{n} = (2(-5) - 1(4), -( (-3)(-5) - 1(1) ), -3(4) - 2(1)) = (-10 - 4, -(15 - 1), -12 - 2) = (-14, -14, -14)$.

   We can simplify this normal vector by dividing by -14 to get $\vec{n} = (1, 1, 1)$.

4. Write the equation of the plane:

   Using the point-normal form of the equation of a plane, we have: $1(x - 0) + 1(y - 7) + 1(z + 7) = 0$, which simplifies to $x + y - 7 + z + 7 = 0$, and further simplifies to $x + y + z = 0$.

Therefore, the equation of the plane is $x + y + z = 0$.