Question

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

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

Hint:

To save time during multiple-choice exams, test the given point directly in the options! Plugging $(0, 7, -7)$ into option (C) gives $0 + 7 + (-7) = 0$, which works perfectly. Checking point $Q(-1, 3, -2)$ from the line gives $-1 + 3 - 2 = 0$, which uniquely isolates option (C) as the correct choice in seconds!

Answer 1

The Correct Option is C

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

Step 2: Key Formula or Approach:
From the equation of the given line, we can identify a point on the line, $Q(-1, 3, -2)$, and its direction vector, $\vec{b} = -3\hat{i} + 2\hat{j} + 1\hat{k}$. Since the plane contains the line, it must contain the point $Q$, and the vector connecting $P$ and $Q$ ($\vec{PQ}$) must lie entirely within the plane. The normal vector ($\vec{n}$) to the plane will be perpendicular to both $\vec{b}$ and $\vec{PQ}$, which can be found using the vector cross product: $$ \vec{n} = \vec{PQ} \times \vec{b} $$

Step 3: Detailed Explanation:
Let's find the components of the vector $\vec{PQ}$ connecting the two points on the plane: $$ \vec{PQ} = (-1 - 0)\hat{i} + (3 - 7)\hat{j} + (-2 - (-7))\hat{k} = -1\hat{i} - 4\hat{j} + 5\hat{k} $$ Now, compute the normal vector $\vec{n}$ by evaluating the determinant cross product of $\vec{PQ}$ and $\vec{b}$: $$ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -4 & 5 \\ -3 & 2 & 1 \end{vmatrix} $$ $$ \vec{n} = \hat{i}[(-4)(1) - (5)(2)] - \hat{j}[(-1)(1) - (5)(-3)] + \hat{k}[(-1)(2) - (-4)(-3)] $$ $$ \vec{n} = \hat{i}(-4 - 10) - \hat{j}(-1 + 15) + \hat{k}(-2 - 12) $$ $$ \vec{n} = -14\hat{i} - 14\hat{j} - 14\hat{k} $$ We can simplify the normal vector components by dividing by the common scalar $-14$, yielding a simplified normal vector $\vec{n'} = 1\hat{i} + 1\hat{j} + 1\hat{k}$. The standard Cartesian equation of a plane passing through $(x_0, y_0, z_0)$ with a normal vector $A\hat{i} + B\hat{j} + C\hat{k}$ is: $$ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 $$ Substituting the point $P(0, 7, -7)$ and our simplified normal coefficients $(1, 1, 1)$: $$ 1(x - 0) + 1(y - 7) + 1(z - (-7)) = 0 $$ $$ x + y - 7 + z + 7 = 0 \implies x + y + z = 0 $$

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