The distance of the point P(1, 2, 1) from the | vedclass

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(A) Let the given line be $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{2} = \lambda$. Any point $A$ on the line is given by $A = (2\lambda + 1, \lambda

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$ \frac{2 \sqrt{5}}{3} $

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(A) Let the given line be $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{2} = \lambda$. Any point $A$ on the line is given by $A = (2\lambda + 1, \lambda + 2, 2\lambda + 3)$. The vector $\vec{PA} = (2\lambda + 1 - 1)\hat{i} + (\lambda + 2 - 2)\hat{j} + (2\lambda + 3 - 1)\hat{k} = 2\lambda\hat{i} + \lambda\hat{j} + (2\lambda + 2)\hat{k}$. The direction vector of the line is $\vec{v} = 2\hat{i} + 1\hat{j} + 2\hat{k}$. Since $\vec{PA}$ is perpendicular to the line,$\vec{PA} \cdot \vec{v} = 0$. $(2\lambda)(2) + (\lambda)(1) + (2\lambda + 2)(2) = 0$. $4\lambda + \lambda + 4\lambda + 4 = 0 \implies 9\lambda = -4 \implies \lambda = -\frac{4}{9}$. Substituting $\lambda = -\frac{4}{9}$ in $A$,we get $A = (2(-\frac{4}{9}) + 1, -\frac{4}{9} + 2, 2(-\frac{4}{9}) + 3) = (\frac{1}{9}, \frac{14}{9}, \frac{19}{9})$. The distance $PA = \sqrt{(\frac{1}{9} - 1)^2 + (\frac{14}{9} - 2)^2 + (\frac{19}{9} - 1)^2} = \sqrt{(-\frac{8}{9})^2 + (-\frac{4}{9})^2 + (\frac{10}{9})^2}$. $PA = \sqrt{\frac{64 + 16 + 100}{81}} = \sqrt{\frac{180}{81}} = \sqrt{\frac{20}{9}} = \frac{2\sqrt{5}}{3}$.

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

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