The length of the projection of the line segm | vedclass

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(C) Let the points be $A(4, -1, 3)$ and $B(5, -1, 4)$. The vector $\overrightarrow{AB} = (5-4)\hat{i} + (-1 - (-1))\hat{j} + (4-3)\hat{k} = \hat{i} +

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$\sqrt{\frac{2}{3}}$

Vedclass · Answer

(C) Let the points be $A(4, -1, 3)$ and $B(5, -1, 4)$. The vector $\overrightarrow{AB} = (5-4)\hat{i} + (-1 - (-1))\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{k}$.The normal to the plane $x + y + z = 7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.The unit normal vector is $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.The projection of $\overrightarrow{AB}$ on the normal $\vec{n}$ is $d = |\overrightarrow{AB} \cdot \hat{n}| = |(\hat{i} + \hat{k}) \cdot \frac{(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}}| = \frac{1+0+1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.The length of the projection of the line segment $AB$ on the plane is given by $\sqrt{|\overrightarrow{AB}|^2 - d^2}$.Here,$|\overrightarrow{AB}|^2 = 1^2 + 0^2 + 1^2 = 2$.So,the length of the projection $= \sqrt{2 - (\frac{2}{\sqrt{3}})^2} = \sqrt{2 - \frac{4}{3}} = \sqrt{\frac{6-4}{3}} = \sqrt{\frac{2}{3}}$.

The length of the projection of the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$ on the plane $x + y + z = 7$ is:

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