Question

The plane passing through the points(2,1,0), (5,0,1)and (4,1,1)intersects the x axis at

• A. (3,0,0)
• B. (-3,0,0)
• C. (0,0,0)
• D. (1,0,0)
• E. (-1,0,0)

Answer 1

The Correct Option is A

Step 1: Find two vectors in the plane

Given points: \( A(2,1,0) \), \( B(5,0,1) \), \( C(4,1,1) \).

Vector \( \overrightarrow{AB} = B - A = \langle 5-2, 0-1, 1-0 \rangle = \langle 3, -1, 1 \rangle \).

Vector \( \overrightarrow{AC} = C - A = \langle 4-2, 1-1, 1-0 \rangle = \langle 2, 0, 1 \rangle \).

Step 2: Find the normal vector to the plane

Compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):

\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ 2 & 0 & 1 \end{vmatrix} \]

\[ = \mathbf{i}((-1)(1)-(1)(0)) - \mathbf{j}((3)(1)-(1)(2)) + \mathbf{k}((3)(0)-(-1)(2)) \]

\[ = -\mathbf{i} - \mathbf{j} + 2\mathbf{k} = \langle -1, -1, 2 \rangle \]

Step 3: Write the plane equation

Using point \( A(2,1,0) \) and normal vector \( \langle -1, -1, 2 \rangle \):

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

Simplify:

\[ -x + 2 - y + 1 + 2z = 0 \]

\[ -x - y + 2z + 3 = 0 \]

\[ x + y - 2z = 3 \]

Step 4: Find x-intercept

For x-intercept, set \( y = z = 0 \):

\[ x + 0 - 0 = 3 \]

\[ x = 3 \]