The projection of the line segment joining P( | vedclass

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(D) Let the points be $P(2, -1, 0)$ and $Q(3, 2, -1)$.The vector $\vec{PQ}$ is given by $(3-2)\hat{i} + (2-(-1))\hat{j} + (-1-0)\hat{k} = 1\hat{i} + 3

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(D) Let the points be $P(2, -1, 0)$ and $Q(3, 2, -1)$.The vector $\vec{PQ}$ is given by $(3-2)\hat{i} + (2-(-1))\hat{j} + (-1-0)\hat{k} = 1\hat{i} + 3\hat{j} - 1\hat{k}$.The direction ratios of the line are $1, 2, 2$. Thus,the direction vector $\vec{v}$ is $1\hat{i} + 2\hat{j} + 2\hat{k}$.The unit vector $\hat{u}$ along the line is $\frac{1\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{1\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{9}} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.The projection of $\vec{PQ}$ on the line is the dot product $\vec{PQ} \cdot \hat{u}$.Projection $= (1\hat{i} + 3\hat{j} - 1\hat{k}) \cdot (\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k})$.Projection $= (1 	imes \frac{1}{3}) + (3 	imes \frac{2}{3}) + (-1 	imes \frac{2}{3}) = \frac{1}{3} + 2 - \frac{2}{3} = \frac{1-2}{3} + 2 = -\frac{1}{3} + 2 = \frac{5}{3}$.Thus,the correct option is $D$.

The projection of the line segment joining $P(2, -1, 0)$ and $Q(3, 2, -1)$ on the line whose direction ratios are $1, 2, 2$ is

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