Question

The distance of the point $P (4,6,-2)$ from the line passing through the point $(-3,2,3)$ and parallel to a line with direction ratios $3,3,-1$ is equal to :

• A. $2 \sqrt{3}$
• B. $\sqrt{14}$
• C. 3
• D. $\sqrt{6}$

Answer 1

The Correct Option is B

To find the distance of a point from a line in 3D space, we use the formula for the distance of a point from a line given by a point and direction ratios. The formula is:

\[ \text{Distance} = \frac{|(x_2 - x_1, y_2 - y_1, z_2 - z_1) \cdot (b_1, b_2, b_3)|}{\sqrt{b_1^2 + b_2^2 + b_3^2}} \]

Where:

• \((x_1, y_1, z_1)\) is a point on the line given as \((-3, 2, 3)\).
• \((x_2, y_2, z_2)\) is the given point \((4, 6, -2)\).
• \((b_1, b_2, b_3)\) are the direction ratios of the line, given as \((3, 3, -1)\).

Substituting the values, we first calculate the vector between the point and the point on the line:

\[ (4 - (-3), 6 - 2, -2 - 3) = (7, 4, -5) \]

Then, find the cross product with the direction ratios:

\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 4 & -5 \\ 3 & 3 & -1 \end{vmatrix} = \hat{i}(4 \cdot (-1) - (-5) \cdot 3) - \hat{j}(7 \cdot (-1) - (-5) \cdot 3) + \hat{k}(7 \cdot 3 - 4 \cdot 3) \]

\[ = \hat{i}(-4 + 15) - \hat{j}(-7 + 15) + \hat{k}(21 - 12) \]

\[ = \hat{i}(11) - \hat{j}(8) + \hat{k}(9) \] \end{div>

Then find the magnitude of this cross product:

\[ \sqrt{11^2 + (-8)^2 + 9^2} = \sqrt{121 + 64 + 81} = \sqrt{266} \]

The magnitude of the direction ratios is:

\[ \sqrt{3^2 + 3^2 + (-1)^2} = \sqrt{9 + 9 + 1} = \sqrt{19} \]

Finally, calculate the distance:

\[ \text{Distance} = \frac{\sqrt{266}}{\sqrt{19}} = \sqrt{14} \]

Thus, the distance of the point \( P(4,6,-2) \) from the line is \( \sqrt{14} \). Therefore, the correct answer is \(\sqrt{14}\).