The square of the distance of the point from the line in the direction of the vector is : [2025]

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Solution

Q.
The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is : [2025]

1 66

2 41

3 44

4 54

Ans.
(1)

Equation of line passing through the point $(\frac{15}{7}, \frac{32}{7}, 7)$ having direction ratios 1, 4 and 7 is given by

$\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7} = λ (say)$

$∴ x = λ + \frac{15}{7}, y = 4 λ + \frac{32}{7} and z = 7 λ + 7$

It lies on the given line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$

i.e., $\frac{λ + \frac{15}{7} + 1}{3} = \frac{7 λ + 7 + 5}{7}$

$\Rightarrow 7 λ + 22 = 21 λ + 36$

$\Rightarrow λ = - 1$

$∴ (x, y, z) \equiv (\frac{8}{7}, \frac{4}{7}, 0)$

Square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ and $(\frac{8}{7}, \frac{4}{7}, 0)$

$= {(\frac{15}{7} - \frac{8}{7})}^{2} + {(\frac{32}{7} - \frac{4}{7})}^{2} + {(7 - 0)}^{2}$

= 1 + 16 + 49 = 66.

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