Question

The equation of the plane through the points (2, 1, 0), (3, 2, -2) and (3, 1, 7) is

• A. 6x - 3y + 2z - 7 = 0
• B. 3x - 2y + 6z - 27 = 0
• C. 7x - 9y - z - 5 = 0
• D. 2x - 3y + 4z - 27 = 0

Hint:

To find the equation of a plane through three points, first find two vectors on the plane by subtracting the coordinates of the points. Then, find the cross product of these vectors to obtain the normal vector to the plane. Finally, use the point-normal form of the plane equation to write the equation of the plane.

Answer 1

The Correct Option is C

The correct answer is: (C): \( 7x - 9y - z - 5 = 0 \)

We are tasked with finding the equation of the plane that passes through the points \( (2, 1, 0) \), \( (3, 2, -2) \), and \( (3, 1, 7) \).

Step 1: Find two vectors on the plane

Let the points be \( P_1(2, 1, 0) \), \( P_2(3, 2, -2) \), and \( P_3(3, 1, 7) \). To find the equation of the plane, we first need two vectors that lie on the plane. We can find these vectors by subtracting the coordinates of one point from the others:

\( \vec{v_1} = P_2 - P_1 = (3 - 2, 2 - 1, -2 - 0) = (1, 1, -2) \)

\( \vec{v_2} = P_3 - P_1 = (3 - 2, 1 - 1, 7 - 0) = (1, 0, 7) \)

Step 2: Find 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 \( \vec{v_2} \):

\( \vec{n} = \vec{v_1} \times \vec{v_2} \)

Now, compute the cross product using the determinant formula:

\( \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & 0 & 7 \end{vmatrix} \)

Step 3: Calculate the cross product

Expanding the determinant, we get:

\( \vec{n} = \hat{i} \begin{vmatrix} 1 & -2 \\ 0 & 7 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ 1 & 7 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \)

Now compute each 2x2 determinant:

\( \vec{n} = \hat{i} (7) - \hat{j} (9) + \hat{k} (-1) \)

\( \vec{n} = 7\hat{i} - 9\hat{j} - \hat{k} \)

Step 4: Use the point-normal form of the plane equation

The equation of the plane can be written as:

\( \vec{n} \cdot (\vec{r} - \vec{P_1}) = 0 \)

Here, \( \vec{r} = (x, y, z) \) is a general point on the plane, and \( \vec{P_1} = (2, 1, 0) \) is a point on the plane. Substituting \( \vec{n} = (7, -9, -1) \) and \( \vec{P_1} = (2, 1, 0) \), we get:

\( 7(x - 2) - 9(y - 1) - (z - 0) = 0 \)

Step 5: Simplify the equation

Expanding and simplifying:

\( 7x - 14 - 9y + 9 - z = 0 \)

\( 7x - 9y - z - 5 = 0 \)

Conclusion:
The equation of the plane is \( 7x - 9y - z - 5 = 0 \), so the correct answer is (C): \( 7x - 9y - z - 5 = 0 \).