Question

The equation of the plane through the point (1, 5, 3) and having a normal vector \(\hat n=2\hat i-2\hat j-\hat k\) is

• A. 2x+2y+z=9
• B. 2x-2y-z=11
• C. 2x+2y-z=9
• D. 2x-2y-z=9
• E. 2x-2y-z=13

Answer 1

The Correct Option is D

The equation of a plane is given by the formula: \[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \] where $(x_1, y_1, z_1)$ is a point on the plane, and $a, b, c$ are the components of the normal vector to the plane.
Here, the normal vector is $\vec{n} = 2\hat{i} - 2\hat{j} - \hat{k}$
So the components of the normal vector are $a = 2$, $b = -2$, and $c = -1$.
The given point is $(1, -5, 3)$. Substituting these into the equation of the plane: \[ 2(x - 1) - 2(y + 5) - (z - 3) = 0 \] Expanding this: \[ 2x - 2 - 2y - 10 - z + 3 = 0 \] Simplifying: \[ 2x - 2y - z - 9 = 0 \] Thus, the equation of the plane is: \[ 2x - 2y - z = 9 \]

The correct option is (D) : \(2x-2y-z=9\)

Answer 2

The equation of the plane through the point (1, 5, 3) and having the normal vector \(\hat{n} = 2\hat{i} - 2\hat{j} - \hat{k}\) is:

We use the equation of the plane \(\hat{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0\), where:

• \(\hat{n} = (2, -2, -1)\) is the normal vector,
• \(\mathbf{r_0} = (1, 5, 3)\) is the given point on the plane,
• \(\mathbf{r} = (x, y, z)\) is a general point on the plane.

The equation becomes:

\(2(x - 1) - 2(y - 5) - (z - 3) = 0\)

Expanding and simplifying:
\[ 2x - 2 - 2y - 10 - z + 3 = 0 \]

\[ 2x - 2y - z - 9 = 0 \]

\[ 2x - 2y - z = 9 \]

Thus, the equation of the plane is:

\[ 2x - 2y - z = 9 \]

Therefore, the correct answer is:

\[ 2x - 2y - z = 9 \]