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

<p>To find the perpendicular distance from the point $P(2,-10,1)$ to the line given by</p> <p>$\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2},$</p> <p>follow these steps:</p> <p><p><strong>Parametrize the Line:</strong></p> <p>From the given symmetric equations, set the common parameter as $t$:</p></p> <p><p>$x = 1 + 2t$</p></p> <p><p>$y = -2 - t$</p></p> <p><p>$z = -3 + 2t$</p> <p>This shows that:</p></p> <p><p>A point on the line is $A(1,-2,-3)$ (when $t=0$).</p></p> <p><p>The direction vector is $\vec{d} = \langle 2, -1, 2 \rangle.$ </p></p> <p><p><strong>Determine the Vector from Point A to P:</strong></p> <p>Calculate</p> <p>$$\vec{AP} = P - A = \langle 2 - 1, \; -10 - (-2), \; 1 - (-3) \rangle = \langle 1, \; -8, \; 4 \rangle.$$</p></p> <p><p><strong>Compute the Cross Product:</strong></p> <p>The formula for the perpendicular distance from a point to a line in 3D is:</p> <p>$d = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}.$</p> <p>First, find the cross product $\vec{AP} \times \vec{d}$:</p> <p>$$ \vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \Big( (-8)(2) - (4)(-1), \; (4)(2) - (1)(2), \; (1)(-1) - (-8)(2) \Big). $$</p> <p>Evaluate each component:</p></p> <p><p>First component: $(-8 \times 2) - (4 \times -1) = -16 + 4 = -12.$</p></p> <p><p>Second component: $(4 \times 2) - (1 \times 2) = 8 - 2 = 6.$</p></p> <p><p>Third component: $(1 \times -1) - (-8 \times 2) = -1 + 16 = 15.$</p> <p>So, </p> <p>$\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle.$</p></p> <p><strong>Calculate the Magnitudes:</strong></p> <p><p>For the cross product:</p> <p>$$ \|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}. $$</p></p> <p><p>For the direction vector:</p> <p>$\|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.$</p></p> <p><p><strong>Compute the Distance:</strong></p> <p>Substitute the magnitudes into the distance formula:</p> <p>$d = \frac{9\sqrt{5}}{3} = 3\sqrt{5}.$</p></p> <p>Thus, the perpendicular distance from the point $P(2,-10,1)$ to the line is $3\sqrt{5}$.</p> <p>Comparing with the options given, the correct answer is:</p> <p>Option C: $3\sqrt{5}$.</p>