The distance of the point A(7, -2, 11) from t | vedclass

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(B) Let the line along which the distance is measured be $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6} = \lambda$.Any point on this line is $P(2\la

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Let the line along which the distance is measured be $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6} = \lambda$.Any point on this line is $P(2\lambda + 7, -3\lambda - 2, 6\lambda + 11)$.Since this point $P$ also lies on the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$,we have:$\frac{2\lambda + 7 - 6}{1} = \frac{-3\lambda - 2 - 4}{0} = \frac{6\lambda + 11 - 8}{3}$From the middle term,the denominator is $0$,so the numerator must be $0$ for the ratio to be defined:$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$.Substituting $\lambda = -2$ into the coordinates of $P$:$x = 2(-2) + 7 = 3$$y = -3(-2) - 2 = 4$$z = 6(-2) + 11 = -1$So,the point of intersection is $B(3, 4, -1)$.The distance $AB$ is the distance between $A(7, -2, 11)$ and $B(3, 4, -1)$:$AB = \sqrt{(7-3)^2 + (-2-4)^2 + (11 - (-1))^2}$$AB = \sqrt{4^2 + (-6)^2 + 12^2}$$AB = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.

The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:

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