The perpendicular distance from the point P(3 | vedclass

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(A) The equation of the line passing through the point $A(2, 1, 0)$ and parallel to the vector $\vec{v} = \hat{i} + 5\hat{j} + 2\hat{k}$ is given by $

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$\frac{1}{\sqrt{6}}$

Vedclass · Answer

(A) The equation of the line passing through the point $A(2, 1, 0)$ and parallel to the vector $\vec{v} = \hat{i} + 5\hat{j} + 2\hat{k}$ is given by $\frac{x-2}{1} = \frac{y-1}{5} = \frac{z-0}{2} = t$. Any point $R$ on this line can be represented as $R(t+2, 5t+1, 2t)$. Let $P(3, 5, 2)$ be the given point. The vector $\vec{PR}$ is given by $\vec{PR} = (t+2-3)\hat{i} + (5t+1-5)\hat{j} + (2t-2)\hat{k} = (t-1)\hat{i} + (5t-4)\hat{j} + (2t-2)\hat{k}$. Since $\vec{PR}$ is perpendicular to the line, its dot product with the direction vector $\vec{v} = \hat{i} + 5\hat{j} + 2\hat{k}$ must be zero: $(t-1)(1) + (5t-4)(5) + (2t-2)(2) = 0$. $t - 1 + 25t - 20 + 4t - 4 = 0$. $30t - 25 = 0 \implies t = \frac{25}{30} = \frac{5}{6}$. Substituting $t = \frac{5}{6}$ into the coordinates of $R$, we get $R = (\frac{5}{6}+2, 5(\frac{5}{6})+1, 2(\frac{5}{6})) = (\frac{17}{6}, \frac{31}{6}, \frac{10}{6})$. The perpendicular distance $d$ is the magnitude of vector $\vec{PR}$: $d = \sqrt{(\frac{17}{6}-3)^2 + (\frac{31}{6}-5)^2 + (\frac{10}{6}-2)^2} = \sqrt{(-\frac{1}{6})^2 + (\frac{1}{6})^2 + (-\frac{2}{6})^2} = \sqrt{\frac{1}{36} + \frac{1}{36} + \frac{4}{36}} = \sqrt{\frac{6}{36}} = \frac{\sqrt{6}}{6} = \frac{1}{\sqrt{6}}$.

The perpendicular distance from the point $P(3, 5, 2)$ to the line $L$ passing through the point $2\hat{i} + \hat{j}$ and parallel to the vector $\hat{i} + 5\hat{j} + 2\hat{k}$ is

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