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

Let the point be $ P(1, 2, -4) $ and the line be $ \frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6} = t $. 

A general point on the line is $ Q(2t+3, 3t+3, 6t-5) $. The vector $ \vec{PQ} $ is:

$$ \vec{PQ} = (2t+3-1, 3t+3-2, 6t-5+4) = (2t+2, 3t+1, 6t-1). $$

The direction vector of the line is $ \vec{d} = (2, 3, 6) $. 

For the shortest distance, $ \vec{PQ} \cdot \vec{d} = 0 $:

$$ (2t+2)(2) + (3t+1)(3) + (6t-1)(6) = 0 \implies 4t+4 + 9t+3 + 36t-6 = 0 \implies 49t+1 = 0 $$

$$\implies t = -\frac{1}{49}. $$

Substitute $ t = -\frac{1}{49} $ into $ Q $:

$$ Q\left(2\left(-\frac{1}{49}\right)+3, 3\left(-\frac{1}{49}\right)+3, 6\left(-\frac{1}{49}\right)-5\right) = \left(\frac{-2+147}{49}, \frac{-3+147}{49}, \frac{-6-245}{49}\right) = \left(\frac{145}{49}, \frac{144}{49}, \frac{-251}{49}\right). $$

The vector $ \vec{PQ} $ is:

$$ \vec{PQ} = \left(\frac{145}{49}-1, \frac{144}{49}-2, \frac{-251}{49}+4\right) = \left(\frac{145-49}{49}, \frac{144-98}{49}, \frac{-251+196}{49}\right) = \left(\frac{96}{49}, \frac{46}{49}, \frac{-55}{49}\right). $$

The distance $ d = |\vec{PQ}| $ is:

$$ d = \frac{1}{49} \sqrt{96^2 + 46^2 + (-55)^2} = \frac{1}{49} \sqrt{9216 + 2116 + 3025} = \frac{1}{49} \sqrt{14357} = \frac{\sqrt{14357}}{49}. $$

Simplify $ \sqrt{14357} $ as $ \sqrt{293 \cdot 49} = 7\sqrt{293} $, so:

$$ d = \frac{7\sqrt{293}}{49} = \frac{\sqrt{293}}{7}. $$

Answer 2

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 \(A = (1, 2, -4)\).

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

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) \(\frac{\sqrt{293}}{7}\).