Question

The distance of the point (1, 2, -4) from the line $\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}$ is

• A. $\frac{293}{7}$
• B. $\frac{\sqrt{293}}{7}$
• C. $\frac{293}{49}$
• D. $\frac{\sqrt{293}}{49}$

Answer 1

The Correct Option is B

We need to find the distance of the point (1, 2, -4) from the line $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$.

The direction vector of the line is $\vec{b} = (2, 3, 6)$. A point on the line is $P = (3, 3, -5)$.

Let the given point be $A = (1, 2, -4)$.

We want to find the distance from point A to the line. This can be calculated using the formula:

$d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|}$

First, find the vector $\overrightarrow{AP} = (3-1, 3-2, -5-(-4)) = (2, 1, -1)$.

Now, find the cross product $\overrightarrow{AP} \times \vec{b}$:

$\overrightarrow{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(1 \cdot 6 - (-1)(3)) - \hat{j}(2 \cdot 6 - (-1)(2)) + \hat{k}(2 \cdot 3 - 1 \cdot 2)$

$\overrightarrow{AP} \times \vec{b} = \hat{i}(6+3) - \hat{j}(12+2) + \hat{k}(6-2) = 9\hat{i} - 14\hat{j} + 4\hat{k} = (9, -14, 4)$

Now, find the magnitude of $\overrightarrow{AP} \times \vec{b}$:

$|\overrightarrow{AP} \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81 + 196 + 16} = \sqrt{293}$

Find the magnitude of $\vec{b}$:

$|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Now, find the distance:

$d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{293}}{7}$

Therefore, the distance of the point from the line is $\frac{\sqrt{293}}{7}$.

Thus, the correct option is (B) $\frac{\sqrt{293}}{7}$.