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 \( y = z = 0 \):

\[ x + 0 - 0 = 3 \]

\[ x = 3 \]

Answer 2

Step 1: Understand the problem and given information.

We are tasked with finding the point where the plane passing through the points \( (2, 1, 0) \), \( (5, 0, 1) \), and \( (4, 1, 1) \) intersects the \( x \)-axis. The general equation of a plane is:

\[ ax + by + cz + d = 0, \]

where \( a, b, c, \) and \( d \) are constants to be determined.

Step 2: Find two direction vectors in the plane.

To find the equation of the plane, we first compute two direction vectors using the given points:

\[ \vec{v}_1 = (5 - 2, 0 - 1, 1 - 0) = (3, -1, 1), \]

\[ \vec{v}_2 = (4 - 2, 1 - 1, 1 - 0) = (2, 0, 1). \]

Step 3: Compute the normal vector to the plane.

The normal vector \( \vec{n} \) to the plane is given by the cross product of \( \vec{v}_1 \) and \( \vec{v}_2 \):

\[ \vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ 2 & 0 & 1 \end{vmatrix}. \]

Expand the determinant:

\[ \vec{n} = \mathbf{i} \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -1 \\ 2 & 0 \end{vmatrix}. \]

Compute each minor determinant:

\[ \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} = (-1)(1) - (1)(0) = -1, \]

\[ \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} = (3)(1) - (1)(2) = 3 - 2 = 1, \]

\[ \begin{vmatrix} 3 & -1 \\ 2 & 0 \end{vmatrix} = (3)(0) - (-1)(2) = 0 + 2 = 2. \]

Substitute back into the cross product:

\[ \vec{n} = -1\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}. \]

Thus, the normal vector is:

\[ \vec{n} = (-1, -1, 2). \]

Step 4: Write the equation of the plane.

The general equation of the plane is:

\[ -1(x - x_0) - 1(y - y_0) + 2(z - z_0) = 0, \]

where \( (x_0, y_0, z_0) \) is a point on the plane. Using \( (2, 1, 0) \):

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

Simplify:

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

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

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

Rewrite:

\[ x + y - 2z = 3. \tag{1} \]

Step 5: Find the intersection with the \( x \)-axis.

On the \( x \)-axis, \( y = 0 \) and \( z = 0 \). Substitute \( y = 0 \) and \( z = 0 \) into equation (1):

\[ x + 0 - 2(0) = 3, \]

\[ x = 3. \]

Final Answer:

The plane intersects the \( x \)-axis at \( (3, 0, 0) \).