A line passes through A(4, -6, -2) and B(16, | vedclass

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(D) The direction vector of line $AB$ is $\vec{v} = (16-4, -2-(-6), 4-(-2)) = (12, 4, 6)$.The unit vector in the direction of $AB$ is $\hat{u} = \frac

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$22$

Vedclass · Answer

(D) The direction vector of line $AB$ is $\vec{v} = (16-4, -2-(-6), 4-(-2)) = (12, 4, 6)$.The unit vector in the direction of $AB$ is $\hat{u} = \frac{(12, 4, 6)}{\sqrt{12^2 + 4^2 + 6^2}} = \frac{(12, 4, 6)}{\sqrt{144 + 16 + 36}} = \frac{(12, 4, 6)}{\sqrt{196}} = \frac{(12, 4, 6)}{14} = (\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$.Point $P$ is at a distance of $21$ units from $A(4, -6, -2)$ along the line $AB$,so $P = A + 21 \hat{u}$.$P = (4, -6, -2) + 21(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}) = (4 + 18, -6 + 6, -2 + 9) = (22, 0, 7)$.Here $a=22, b=0, c=7$,which are non-negative integers.The distance between $P(22, 0, 7)$ and $Q(4, -12, 3)$ is $\sqrt{(22-4)^2 + (0-(-12))^2 + (7-3)^2}$.$= \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$.

$A$ line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$,where $a, b, c$ are non-negative integers,lies on the line $AB$ at a distance of $21$ units from point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to...........

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