The projection of the line segment joining th | vedclass

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(A) Let the points be $P(2,1,-3)$ and $Q(-1,0,2)$.The vector $\vec{PQ} = (-1-2)\hat{i} + (0-1)\hat{j} + (2-(-3))\hat{k} = -3\hat{i} - 1\hat{j} + 5\hat

Vedclass · Accepted Answer

$\frac{19}{7}$ units

Vedclass · Answer

(A) Let the points be $P(2,1,-3)$ and $Q(-1,0,2)$.The vector $\vec{PQ} = (-1-2)\hat{i} + (0-1)\hat{j} + (2-(-3))\hat{k} = -3\hat{i} - 1\hat{j} + 5\hat{k}$.The direction ratios of the line are $3, 2, 6$. The direction vector of the line is $\vec{v} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.The magnitude of the direction vector is $|\vec{v}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.The unit vector along the line is $\hat{u} = \frac{3\hat{i} + 2\hat{j} + 6\hat{k}}{7}$.The projection of the line segment $PQ$ on the line is given by the absolute value of the dot product of $\vec{PQ}$ and $\hat{u}$:Projection $= |\vec{PQ} \cdot \hat{u}| = |(-3\hat{i} - 1\hat{j} + 5\hat{k}) \cdot \frac{(3\hat{i} + 2\hat{j} + 6\hat{k})}{7}|$$= |\frac{(-3)(3) + (-1)(2) + (5)(6)}{7}| = |\frac{-9 - 2 + 30}{7}| = |\frac{19}{7}| = \frac{19}{7}$ units.

The projection of the line segment joining the points $(2,1,-3)$ and $(-1,0,2)$ on the line whose direction ratios are $3,2,6$ is

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