Step 1: Understanding the Concept:
We need to calculate the perpendicular distance from a given point $P$ to a line $L$ in 3D space. This can be efficiently computed using vector algebra by finding the magnitude of the cross product of the vector connecting a point on the line to $P$ and the line's direction vector.
Step 2: Key Formula or Approach:
The perpendicular distance $d$ from a point $P$ to a line passing through point $A$ with direction vector $\bar{b}$ is:
\[ d = \frac{|\vec{AP} \times \bar{b}|}{|\bar{b}|} \]
where $\vec{AP}$ is the vector from point $A$ on the line to the given point $P$.
Step 3: Detailed Explanation:
Let the given point be $P(-3, 2, 3)$.
The line passes through the point $A(4, 6, -2)$.
The direction vector of the line is $\bar{b} = -1\hat{i} + 2\hat{j} + 3\hat{k} = \langle -1, 2, 3 \rangle$.
First, construct the vector $\vec{AP}$:
\[ \vec{AP} = \text{Position vector of } P - \text{Position vector of } A \]
\[ \vec{AP} = \langle -3 - 4, 2 - 6, 3 - (-2) \rangle = \langle -7, -4, 5 \rangle \]
\[ \vec{AP} = -7\hat{i} - 4\hat{j} + 5\hat{k} \]
Next, compute the cross product $\vec{AP} \times \bar{b}$:
\[ \vec{AP} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-7 & -4 & 5
-1 & 2 & 3 \end{vmatrix} \]
\[ = \hat{i}((-4)(3) - (5)(2)) - \hat{j}((-7)(3) - (5)(-1)) + \hat{k}((-7)(2) - (-4)(-1)) \]
\[ = \hat{i}(-12 - 10) - \hat{j}(-21 + 5) + \hat{k}(-14 - 4) \]
\[ = -22\hat{i} - \hat{j}(-16) - 18\hat{k} = \langle -22, 16, -18 \rangle \]
Now, find the magnitude of this cross product vector:
\[ |\vec{AP} \times \bar{b}| = \sqrt{(-22)^2 + 16^2 + (-18)^2} \]
\[ |\vec{AP} \times \bar{b}| = \sqrt{484 + 256 + 324} = \sqrt{1064} \]
Next, find the magnitude of the direction vector $\bar{b}$:
\[ |\bar{b}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
Finally, calculate the perpendicular distance $d$:
\[ d = \frac{|\vec{AP} \times \bar{b}|}{|\bar{b}|} = \frac{\sqrt{1064}}{\sqrt{14}} = \sqrt{\frac{1064}{14}} \]
\[ d = \sqrt{76} \]
Simplify the radical:
\[ d = \sqrt{4 \times 19} = 2\sqrt{19} \]
Step 4: Final Answer:
The perpendicular distance is $2\sqrt{19}$ units.