The length of the perpendicular from the point $(2,-1,4)$ on the straight line, $\frac{x+3}{10}=\fra

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The length of the perpendicular from the point $(2,-1,4)$ on the straight line, $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ is\

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(b) : Foot of perpendicular from a point on a line is $\vec{a}+\frac{\vec{b} \cdot(\vec{c}-\vec{a}) \vec{b}}{\vec{b} \cdot \vec{b}}$. $$ \left\{\begin{array}{l} P \ \frac{(2,-1,4)}{10}=\frac{y-2}{-7}=\frac{z}{1} \end{array}ight. $$ Where, $\vec{a}=-3 \hat{i}+2 \hat{j}$, $$ \vec{b}=10 \hat{i}-7 \hat{j}+\hat{k}, \vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} $$ ∴ Foot of perpendicular\ $=\vec{a}+\frac{(10 \hat{i}-7 \hat{j}+\hat{k}) \cdot(5 \hat{i}-3 \hat{j}+4 \hat{k}) \vec{b}}{150}$\ $=\vec{a}+\frac{(50+21+4) \vec{b}}{150}=\vec{a}+\frac{75}{150} \vec{b}=\vec{a}+\frac{\vec{b}}{2}$\ $=(-3 \hat{i}+2 \hat{j})+\frac{1}{2}(10 \hat{i}-7 \hat{j}+\hat{k})=2 \hat{i}-\frac{3}{2} \hat{j}+\frac{1}{2} \hat{k}$\ ∴ Length of perpendicular from $(2,-1,4)$ to $\left(2,-\frac{3}{2}, \frac{1}{2}ight)$\ $=\sqrt{(2-2)^{2}+\left(-1+\frac{3}{2}ight)^{2}+\left(4-\frac{1}{2}ight)^{2}}$\ $=\sqrt{\frac{1}{4}+\frac{49}{4}}=\sqrt{\frac{50}{4}}=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}}=3.54$\ Thus, length of perpendicular is greater than 3 but less than 4 .\