Question

The perpendicular distance of the line \( \frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z + 3}{2} \) from the point \( P(2, -10, 1) \) is:

• A. \( 3\sqrt{5} \)
• B. \( 5\sqrt{2} \)
• C. \( 4\sqrt{3} \)
• D. \( 6 \)

Hint:

For finding the perpendicular distance from a point to a line in three dimensions: - Convert the given symmetric form of the line to its direction ratios. - Use the distance formula \( d = \frac{| \mathbf{a} \cdot (\mathbf{r_0} - \mathbf{r_1}) |}{|\mathbf{a}|} \) where \( \mathbf{a} \) is the direction vector of the line.

Answer 1

The Correct Option is A

To find the perpendicular distance of the line from the point \( P(2, -10, 1) \), we first express the line's parametric equations. The given symmetric form is:

\(\frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z + 3}{2} = t\)

From this, we derive the parametric equations:

• \(x = 2t + 1\)
• \(y = -t - 2\)
• \(z = 2t - 3\)

The line's direction vector is \(\mathbf{d} = \langle 2, -1, 2 \rangle\) and a point on the line is \(A(1, -2, -3)\).

The vector from point \(P(2, -10, 1)\) to point \(A(1, -2, -3)\) on the line is:

\(\mathbf{AP} = \langle 2-1, -10+2, 1+3 \rangle = \langle 1, -8, 4 \rangle\)

The formula for the perpendicular distance \(d\) from a point to a line given the direction vector \(\mathbf{d}\) and point vector \(\mathbf{AP}\) is:

\(d = \frac{\|\mathbf{AP} \times \mathbf{d}\|}{\|\mathbf{d}\|}\)

Calculating \(\mathbf{AP} \times \mathbf{d}\):

\(\mathbf{AP} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \mathbf{i}( (-8)(2)-(4)(-1) ) - \mathbf{j}( (1)(2)-(4)(2) ) + \mathbf{k}( (1)(-1)+8(2) )\)

\(= \mathbf{i}(-16 + 4) - \mathbf{j}(2 - 8) + \mathbf{k}(-1 + 16)\)

\(= \langle -12, 6, 15 \rangle\)

Now calculate its magnitude:

\(\|\mathbf{AP} \times \mathbf{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405}\)

\(\|\mathbf{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)

The perpendicular distance \(d\) is:

\(d = \frac{\sqrt{405}}{3} = \frac{\sqrt{9 \times 45}}{3} = \frac{3\sqrt{45}}{3} = \sqrt{45} = 3\sqrt{5}\)

Answer 2

The line is given by $$\frac{x-1}{2} = \frac{y+2}{-1} = \frac{z+3}{2}.$$ Let the line be in the form: $$\vec{r} = \vec{a} + \lambda \vec{b},$$ where $$\vec{a} = (1,-2,-3),\quad \vec{b} = (2,-1,2).$$

The given point is $$P(2,-10,1).$$ The perpendicular distance of a point from a line in 3D is given by $$D = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|},$$ where $\overrightarrow{AP} = \vec{P}-\vec{A}$.

Compute $$\overrightarrow{AP} = (2-1,\,-10+2,\,1+3) = (1,-8,4).$$

Now find the cross product $\overrightarrow{AP} \times \vec{b}$: $$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -8 & 4\\ 2 & -1 & 2 \end{vmatrix} = \mathbf{i}((-8)(2) - 4(-1)) - \mathbf{j}((1)(2) - 4(2)) + \mathbf{k}((1)(-1) - (-8)(2)). $$

Simplify: $$ = \mathbf{i}(-16 + 4) - \mathbf{j}(2 - 8) + \mathbf{k}(-1 + 16) = \mathbf{i}(-12) - \mathbf{j}(-6) + \mathbf{k}(15) = (-12, 6, 15). $$

So, $$|\overrightarrow{AP} \times \vec{b}| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}.$$

Also, $$|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3.$$

Therefore, the perpendicular distance is $$D = \frac{9\sqrt{5}}{3} = 3\sqrt{5}.$$

Final Answer: The perpendicular distance of the line from the point is $$\boxed{3\sqrt{5}}.$$