If the direction ratios of a line are proport | vedclass

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(C) Let the points be $A(5, 2, 3)$ and $B(-1, 0, 2)$.The vector $\vec{AB}$ is given by $(-1-5)\hat{i} + (0-2)\hat{j} + (2-3)\hat{k} = -6\hat{i} - 2\ha

Vedclass · Accepted Answer

$13/\sqrt{14}$

Vedclass · Answer

(C) Let the points be $A(5, 2, 3)$ and $B(-1, 0, 2)$.The vector $\vec{AB}$ is given by $(-1-5)\hat{i} + (0-2)\hat{j} + (2-3)\hat{k} = -6\hat{i} - 2\hat{j} - \hat{k}$.The direction ratios of the line are $1, 2, 3$,so the direction vector of the line is $\vec{v} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.The unit vector along the line is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{14}}$.The projection of the line segment $AB$ on the line is the absolute value of the dot product of $\vec{AB}$ and $\hat{u}$.Projection $= |\vec{AB} \cdot \hat{u}| = |(-6\hat{i} - 2\hat{j} - \hat{k}) \cdot \frac{\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{14}}|$$= |\frac{(-6)(1) + (-2)(2) + (-1)(3)}{\sqrt{14}}| = |\frac{-6 - 4 - 3}{\sqrt{14}}| = |\frac{-13}{\sqrt{14}}| = \frac{13}{\sqrt{14}}$.

If the direction ratios of a line are proportional to $1, 2, 3$,find the projection of the line segment joining the points $(5, 2, 3)$ and $(-1, 0, 2)$ on the line.

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