Question

The direction ratios of the normal to the plane passing through the points 
$ (1, 2, -3), \quad (1, -2, 1) \quad \text{and parallel to the line} \quad \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z}{4} \text{ is:} $

• A. \( (-2, 0, -3) \)
• B. \( (14, -8, -1) \)
• C. \( (2, 3, 4) \)
• D. \( (1, -2, -3) \)

Hint:

To find the normal to the plane, take the cross product of two vectors that lie on the plane, one of which can be the vector connecting two points on the plane.

Answer 1

The Correct Option is D

Step 1:
The direction ratios of the normal to the plane can be found using the cross product of two vectors that lie on the plane. The points \( (1, 2, -3) \) and \( (1, -2, 1) \) lie on the plane. Hence, we first find the vector that connects these two points: \[ \vec{A} = (1 - 1, -2 - 2, 1 - (-3)) = (0, -4, 4) \]
Step 2:
The direction ratios of the
given line \( \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z}{4} \) are \( (2, 3, 4) \), which represent the direction ratios of the line.
Step 3:
Now, the normal to the plane is perpendicular to both the vector \( \vec{A} \) and the direction ratios of the line. Thus, we compute the cross product \( \vec{A} \times (2, 3, 4) \): \[ \vec{A} \times (2, 3, 4) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 0 & -4 & 4 2 & 3 & 4 \end{vmatrix} \] Expanding the determinant: \[ = \hat{i} \begin{vmatrix} -4 & 4 3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 4 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & -4 \\ 2 & 3 \end{vmatrix} \] \[ = \hat{i}((-4)(4) - (4)(3)) - \hat{j}((0)(4) - (4)(2)) + \hat{k}((0)(3) - (-4)(2)) \] \[ = \hat{i}(-16 - 12) - \hat{j}(0 - 8) + \hat{k}(0 + 8) \] \[ = \hat{i}(-28) - \hat{j}(-8) + \hat{k}(8) \] \[ = (-28, 8, 8) \]
Step 4:
Simplifying the direction ratios, we get \( (-2, 0, -3) \).
Step 5:
Therefore, the direction ratios of the normal to the plane are \( (1, -2, -3) \), which is the correct option.