The distance of the point B(1, 2, 3) from the | vedclass

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(B) Let the line pass through $A(4, 2, 2)$ and be parallel to $\vec{c} = 2i + 3j + 6k$.Let $M$ be the foot of the perpendicular from $B(1, 2, 3)$ to t

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$\sqrt{10}$

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(B) Let the line pass through $A(4, 2, 2)$ and be parallel to $\vec{c} = 2i + 3j + 6k$.Let $M$ be the foot of the perpendicular from $B(1, 2, 3)$ to the line.Vector $\vec{AB} = (1-4)i + (2-2)j + (3-2)k = -3i + 0j + k$.The length $AM$ is the projection of $\vec{AB}$ on the vector $\vec{c}$.$AM = \frac{|\vec{AB} \cdot \vec{c}|}{|\vec{c}|} = \frac{|(-3)(2) + (0)(3) + (1)(6)|}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{|-6 + 0 + 6|}{\sqrt{4 + 9 + 36}} = \frac{0}{7} = 0$.In the right-angled triangle $	riangle ABM$,$BM^2 = AB^2 - AM^2$.$AB^2 = |\vec{AB}|^2 = (-3)^2 + 0^2 + 1^2 = 9 + 0 + 1 = 10$.$BM^2 = 10 - 0^2 = 10$.Therefore,$BM = \sqrt{10}$.

The distance of the point $B(1, 2, 3)$ from the line passing through $A(4, 2, 2)$ and parallel to the vector $\vec{c} = 2i + 3j + 6k$ is

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