Question

The length of the perpendicular from the point (0, 2, 3) on the line $\frac{x+3}{5} = \frac{y-1}{2} =$ The equ

• A. $\sqrt{\frac{59}{30}}$

Answer 1

Answer: (A)

$\sqrt{\frac{59}{30}}$

Answer 2

We assume that the full equation of the line is

$$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z-1}{1}\,,$$

so that a point on the line is

$$Q(-3,\,1,\,1)$$

and its direction vector is

$$\vec{d}=(5,\,2,\,1)\,.$$

Let the given point be

$$P(0,\,2,\,3)\,.$$

Then the vector from $Q$ to $P$ is

$$\vec{QP}=P-Q=(0-(-3),\,2-1,\,3-1)=(3,\,1,\,2)\,.$$

The formula for the distance from a point to a line in 3D is

$$\text{Distance} = \frac{\|\vec{QP}\times\vec{d}\|}{\|\vec{d}\|}\,.$$

First, compute the cross product:

$$\vec{QP}\times\vec{d}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\[4mm] 3&1&2\\[2mm] 5&2&1 \end{vmatrix} =\Bigl(1\cdot1-2\cdot2,\; 2\cdot5-3\cdot1,\; 3\cdot2-1\cdot5\Bigr) =(-3,\;7,\;1)\,.$$

Its magnitude is

$$\|\vec{QP}\times\vec{d}\|=\sqrt{(-3)^2+7^2+1^2}=\sqrt{9+49+1}=\sqrt{59}\,.$$

Next, compute the magnitude of $\vec{d}$:

$$\|\vec{d}\|=\sqrt{5^2+2^2+1^2}=\sqrt{25+4+1}=\sqrt{30}\,.$$

Thus, the perpendicular distance is

$$\text{Distance}=\frac{\sqrt{59}}{\sqrt{30}}=\sqrt{\frac{59}{30}}\,.$$