Question

The plane containing the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ is

• A. $x - y + z = 1$
• B. $x + y + z = 5$
• C. $x + 2y - z = 1$
• D. $2x - y + z = 5$

Answer 1

The Correct Option is A

1. Identify the point and line:

The plane must contain the point \( (3, 2, 0) \) and the line \( \frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4} \).

2. Find two points on the line:

For \( t = 0 \): \( (3, 6, 4) \) (given).

For \( t = 1 \): \( (4, 11, 8) \).

3. Compute two vectors in the plane:

Vector from \( (3, 2, 0) \) to \( (3, 6, 4) \): \( \vec{v_1} = (0, 4, 4) \).

Vector from \( (3, 2, 0) \) to \( (4, 11, 8) \): \( \vec{v_2} = (1, 9, 8) \).

4. Find the normal vector to the plane:

Compute the cross product \( \vec{v_1} \times \vec{v_2} \):

\[ \vec{v_1} \times \vec{v_2} = (4 \cdot 8 - 4 \cdot 9, -(0 \cdot 8 - 4 \cdot 1), 0 \cdot 9 - 4 \cdot 1) = (-4, 4, -4) \]

Simplify to \( (1, -1, 1) \).

5. Write the plane equation:

Using point \( (3, 2, 0) \) and normal vector \( (1, -1, 1) \):

\[ 1(x - 3) - 1(y - 2) + 1(z - 0) = 0 \implies x - y + z = 1 \]

Correct Answer: (A) \( x - y + z = 1 \)

Answer 2

The given line is expressed in symmetric form:

$$ \frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4} $$

The direction vector of this line is $ \mathbf{v} = \langle 1, 5, 4 \rangle $.

• We are given one point on the plane: $ P = (3, 2, 0) $.
• We need another point on the plane. Let's find a point on the given line. If we set $ x = 3 $ (the $ x $-coordinate of the given point), we get:

$$ \frac{3 - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4} = 0 $$

This gives us the point $ Q = (3, 6, 4) $.

• Vector 1: The vector from $ P $ to $ Q $ is $ \mathbf{PQ} = Q - P = \langle 3-3, 6-2, 4-0 \rangle = \langle 0, 4, 4 \rangle $.
• Vector 2: The direction vector of the line, $ \mathbf{v} = \langle 1, 5, 4 \rangle $, lies in the plane.

The normal vector to the plane is perpendicular to both $ \mathbf{PQ} $ and $ \mathbf{v} $. We can find this using the cross product:

$$ \mathbf{n} = \mathbf{PQ} \times \mathbf{v} = \langle 0, 4, 4 \rangle \times \langle 1, 5, 4 \rangle $$ $$ \mathbf{n} = \langle (4 \cdot 4 - 4 \cdot 5), (4 \cdot 1 - 0 \cdot 4), (0 \cdot 5 - 4 \cdot 1) \rangle = \langle -4, 4, -4 \rangle $$

We can simplify this normal vector by dividing by $-4$: $ \mathbf{n} = \langle 1, -1, 1 \rangle $.

The equation of a plane is given by:

$$ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 $$

Where $ (x_0, y_0, z_0) $ is a point on the plane, and $ \langle A, B, C \rangle $ is the normal vector. Using point $ P(3, 2, 0) $ and the normal vector $ \langle 1, -1, 1 \rangle $:

$$ 1(x - 3) - 1(y - 2) + 1(z - 0) = 0 $$ $$ x - 3 - y + 2 + z = 0 $$ $$ x - y + z = 1 $$

Therefore, the equation of the plane is $ x - y + z = 1 $.