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. $x + 3y + z = 14$

Hint:

Save time by testing the given point $(0, 7, -7)$ in the options!
Option (A): $2(0) + 7 - 7 = 0$ (Valid)
Option (B): $0 + 7 - 7 = 0$ (Valid)
Now test the point from the line $(-1, 3, -2)$ to break the tie:
For (A): $2(-1) + 3 - 2 = -1 \neq 0$
For (B): $-1 + 3 - 2 = 0$ (Valid!)
This single trick identifies the correct option in seconds without computing cross products.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the Cartesian equation of a plane that completely contains a given line and also passes through an external point $(0, 7, -7)$.

Step 2: Key Formula or Approach:
From the line equation $\frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$, we can extract two vital pieces of geometric data:
1. A point on the line: $A(-1, 3, -2)$.
2. The directional ratios of the line: $\bar{u} = -3\hat{i} + 2\hat{j} + \hat{k}$.
The plane also passes through the given point $B(0, 7, -7)$. Therefore, a vector connecting the two points on the plane is $\overline{AB} = (0 - (-1))\hat{i} + (7 - 3)\hat{j} + (-7 - (-2))\hat{k} = \hat{i} + 4\hat{j} - 5\hat{k}$.
The normal vector $\bar{n}$ to the plane must be perpendicular to both $\bar{u}$ and $\overline{AB}$, so it can be calculated via their cross product: $\bar{n} = \bar{u} \times \overline{AB}$.

Step 3: Detailed Explanation:
Let's find the normal vector using a standard $3 \times 3$ determinant:
$$\bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5 \end{vmatrix}$$ Expanding along the first row:
$$\bar{n} = \hat{i}\big((2)(-5) - (1)(4)\big) - \hat{j}\big((-3)(-5) - (1)(1)\big) + \hat{k}\big((-3)(4) - (2)(1)\big)$$ $$\bar{n} = \hat{i}(-10 - 4) - \hat{j}(15 - 1) + \hat{k}(-12 - 2)$$ $$\bar{n} = -14\hat{i} - 14\hat{j} - 14\hat{k}$$ We can divide this normal vector by $-14$ to simplify the coefficients to a clean directional ratio: $\bar{n}' = \hat{i} + \hat{j} + \hat{k}$. Thus, $A = 1, B = 1, C = 1$.
Now, form the equation of the plane using point $B(0,7,-7)$:
$$1(x - 0) + 1(y - 7) + 1(z - (-7)) = 0$$ $$x + y - 7 + z + 7 = 0$$ $$x + y + z = 0$$

Step 4: Final Answer:
The equation of the plane is $x + y + z = 0$, which matches option (B).