Question

The equation of the plane passing through $(-1, 5, -7)$ and parallel to the plane $2x - 5y + 7z + 11 = 0$, is:

• A. $\vec{r} \cdot (2\hat{i} - 5\hat{j} - 7\hat{k}) + 76 = 0$
• B. $\vec{r} \cdot (2\hat{i} - 5\hat{j} + 7\hat{k}) + 76 = 0$
• C. $\vec{r} \cdot (2\hat{i} - 5\hat{j} + 7\hat{k}) + 75 = 0$
• D. $\vec{r} \cdot (2\hat{i} - 5\hat{j} + 7\hat{k}) + 65 = 0$
• E. $\vec{r} \cdot (2\hat{i} - 5\hat{j} - 7\hat{k}) + 55 = 0$

Hint:

To quickly find the constant, calculate the dot product of the normal vector $(2, -5, 7)$ and the point $(-1, 5, -7)$. The result is $-76$. Since the plane equation is $Ax+By+Cz = \text{constant}$, you get $2x - 5y + 7z = -76$, which simplifies to $+ 76 = 0$.

Answer 1

The Correct Option is B

Concept: Parallel planes share the same normal vector. The equation of a plane parallel to $Ax + By + Cz + D = 0$ is of the form $Ax + By + Cz + k = 0$. The constant $k$ is determined by substituting the coordinates of the point through which the plane passes.

Step 1: Identify the normal vector and the general form.
The given plane is $2x - 5y + 7z + 11 = 0$. The normal vector is $\vec{n} = 2\hat{i} - 5\hat{j} + 7\hat{k}$.
Any parallel plane will have the equation: \[ 2x - 5y + 7z + k = 0 \]

Step 2: Substitute the given point to find $k$.
The plane passes through $(-1, 5, -7)$. Substitute $x = -1$, $y = 5$, and $z = -7$:
\[ 2(-1) - 5(5) + 7(-7) + k = 0 \] \[ -2 - 25 - 49 + k = 0 \] \[ -76 + k = 0 \quad \Rightarrow \quad k = 76 \]

Step 3: Write the equation in vector form.
The Cartesian equation is $2x - 5y + 7z + 76 = 0$. In vector form $\vec{r} \cdot \vec{n} + d = 0$: \[ \vec{r} \cdot (2\hat{i} - 5\hat{j} + 7\hat{k}) + 76 = 0 \]