The length of the projection of the line segm | vedclass

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(A) Let $A = (5, -1, 4)$ and $B = (4, -1, 3)$.The vector $\vec{AB} = (4-5)\hat{i} + (-1-(-1))\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{k}$.The magnitud

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

Vedclass · Answer

(A) Let $A = (5, -1, 4)$ and $B = (4, -1, 3)$.The vector $\vec{AB} = (4-5)\hat{i} + (-1-(-1))\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{k}$.The magnitude of the vector is $|\vec{AB}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2}$.The normal vector to the plane $x+y+z=7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.Let $	heta$ be the angle between the line segment $AB$ and the plane. The angle $\phi$ between the line segment and the normal to the plane is given by $\cos \phi = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{AB}| |\vec{n}|}$.$\cos \phi = \frac{|(-1)(1) + (0)(1) + (-1)(1)|}{\sqrt{2} \sqrt{1^2+1^2+1^2}} = \frac{|-2|}{\sqrt{2} \sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.Since $\phi$ is the angle with the normal,the angle $	heta$ with the plane is $90^\circ - \phi$,so $\sin 	heta = \cos \phi = \sqrt{\frac{2}{3}}$.Then $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}}$.The length of the projection of the line segment on the plane is $|\vec{AB}| \cos 	heta = \sqrt{2} 	imes \sqrt{\frac{1}{3}} = \sqrt{\frac{2}{3}}$ units.

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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