The distance of the point (1, 2, -4) from the | vedclass

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(B) Let the given point be $A(1, 2, -4)$ and the line be $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6} = \lambda$.Any point $P$ on the line is given

Vedclass · Accepted Answer

$\frac{\sqrt{293}}{7}$

Vedclass · Answer

(B) Let the given point be $A(1, 2, -4)$ and the line be $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6} = \lambda$.Any point $P$ on the line is given by $(2\lambda+3, 3\lambda+3, 6\lambda-5)$.The direction ratios of the line $AP$ are $(2\lambda+3-1, 3\lambda+3-2, 6\lambda-5-(-4))$,which simplifies to $(2\lambda+2, 3\lambda+1, 6\lambda-1)$.Since $AP$ is perpendicular to the given line with direction ratios $(2, 3, 6)$,their dot product must be zero:$2(2\lambda+2) + 3(3\lambda+1) + 6(6\lambda-1) = 0$$4\lambda + 4 + 9\lambda + 3 + 36\lambda - 6 = 0$$49\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{49}$.The square of the distance $AP$ is $AP^2 = (2\lambda+2)^2 + (3\lambda+1)^2 + (6\lambda-1)^2$.Substituting $\lambda = -\frac{1}{49}$:$AP^2 = 49\lambda^2 + 2\lambda + 6 = 49(-\frac{1}{49})^2 + 2(-\frac{1}{49}) + 6$$AP^2 = \frac{1}{49} - \frac{2}{49} + 6 = 6 - \frac{1}{49} = \frac{294-1}{49} = \frac{293}{49}$.Therefore,$AP = \sqrt{\frac{293}{49}} = \frac{\sqrt{293}}{7}$.

The distance of the point $(1, 2, -4)$ from the line $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is

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