Question

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

• A. 2x + y - z = 14
• B. x + y + z = 0
• C. x + 2y + z = 7
• D. 2x + y + z = 0

Answer 1

Answer: (B)

x + y + z = 0

Answer 2

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

   The line in symmetric form is

   $$\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1} = t.$$

   Thus, the parametric equations are:

   $$x = -1 - 3t,\quad y = 3 + 2t,\quad z = -2 + t.$$

   When $t = 0$, a point on the line is $Q(-1,\,3,\,-2)$. The direction vector of the line is:

   $$\vec{d} = (-3,\,2,\,1).$$

2. Determine a second vector in the plane:

   The plane passes through $P(0,\,7,\,-7)$ and contains the line. So, the vector from $Q(-1,\,3,\,-2)$ to $P(0,\,7,\,-7)$ is:

   $$\vec{QP} = (0 - (-1),\, 7 - 3,\, -7 - (-2)) = (1,\,4,\,-5).$$

3. Find the normal vector ($\vec{n}$) to the plane:

   The normal vector is the cross product of $\vec{d}$ and $\vec{QP}$:

   $$\vec{n} = \vec{d} \times \vec{QP} = (-3,\,2,\,1) \times (1,\,4,\,-5).$$

   Computing the cross product:

   $$\vec{n} = \Big(2(-5) - 1(4),\; -\big((-3)(-5) - 1(1)\big),\; -3(4) - 2(1)\Big)$$ $$= \Big(-10 - 4,\; -\big(15 - 1\big),\; -12 - 2\Big)$$ $$= (-14,\,-14,\,-14).$$

   Dividing by $-14$ (since the equation remains unchanged if the normal vector is multiplied by a scalar), we get:

   $$\vec{n} = (1,\,1,\,1).$$

4. Write the equation of the plane:

   The plane passing through $P(0,\,7,\,-7)$ with normal vector $(1,\,1,\,1)$ is given by:

   $$1(x-0) + 1(y-7) + 1(z+7) = 0.$$

   Simplify:

   $$x + y - 7 + z + 7 = 0 \quad \Rightarrow \quad x + y + z = 0.$$

Answer:
Option b) $x+y+z = 0$

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Explanation (Core):
Find a point on the line $Q(-1, 3, -2)$ and direction vector $\vec{d}=(-3,2,1)$. Compute vector from $Q$ to $P(0,7,-7)$: $(1,4,-5)$. Find the normal using the cross product: $\vec{n} = (1,1,1)$. Using the point-normal form, the plane equation is $x+y+z=0$.