Question

Let \( \bar{n} \) be a unit vector normal to the plane \( \pi \) containing the vectors \( \bar{T} + 3\bar{k} \) and \( 2\bar{i} + \bar{j} - \bar{k} \). If this plane \( \pi \) passes through the point \( (-3,7,1) \) and \( p \) is the perpendicular distance from the origin to this plane, then \( \sqrt{p^2 + 5} \) is:

• A. \( 59 \)
• B. \( 8 \)
• C. \( 64 \)
• D. \( 51 \)

Hint:

To find the equation of a plane, use the cross product to determine the normal vector. The perpendicular distance formula helps in evaluating the shortest distance from a point to the plane.

Answer 1

The Correct Option is B

Step 1: Finding the Normal Vector to the Plane
The given vectors lying in the plane are:
\[ \bar{a} = \bar{i} + 3\bar{k}, \quad \bar{b} = 2\bar{i} + \bar{j} - \bar{k}. \] The normal vector to the plane is given by the cross product:
\[ \bar{n} = \bar{a} \times \bar{b}. \] Computing the determinant:
\[ \bar{n} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & 0 & 3 \\ 2 & 1 & -1 \end{vmatrix} \] Expanding along the first row:
\[ \bar{n} = \bar{i} \begin{vmatrix} 0 & 3 \\ 1 & -1 \end{vmatrix} - \bar{j} \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} + \bar{k} \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix}. \] Evaluating the determinants:
\[ \begin{vmatrix} 0 & 3 \\ 1 & -1 \end{vmatrix} = (0 \times -1) - (3 \times 1) = -3, \] \[ \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} = (1 \times -1) - (3 \times 2) = -1 - 6 = -7, \] \[ \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = (1 \times 1) - (0 \times 2) = 1. \] Thus, the normal vector is:
\[ \bar{n} = (-3)\bar{i} + (7)\bar{j} + (1)\bar{k}. \]

Step 2: Equation of the Plane
The plane equation is given by:
\[ -3(x + 3) + 7(y - 7) + 1(z - 1) = 0. \] Simplifying,
\[ -3x - 9 + 7y - 49 + z - 1 = 0. \] \[ -3x + 7y + z - 59 = 0. \]

Step 3: Finding Perpendicular Distance from Origin
The perpendicular distance from the origin \( (0,0,0) \) to the plane is given by:
\[ p = \frac{| -3(0) + 7(0) + 1(0) - 59 |}{\sqrt{(-3)^2 + 7^2 + 1^2}}. \] \[ p = \frac{| -59 |}{\sqrt{9 + 49 + 1}} = \frac{59}{\sqrt{59}} = \sqrt{59}. \]

Step 4: Finding \( \sqrt{p^2 + 5} \)
\[ \sqrt{p^2 + 5} = \sqrt{59 + 5} = \sqrt{64} = 8. \] Thus, the final answer is:
\[ \boxed{8}. \]