The angle between the line r = (i + 2j - k) + | vedclass

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Question

(A) The line is given by $r = (i + 2j - k) + \lambda (i - j + k)$. The direction vector of the line is $b = i - j + k$.The plane is given by $r \cdot

Vedclass · Accepted Answer

$\sin^{-1} \left( \frac{2\sqrt{2}}{3} \right)$

Vedclass · Answer

(A) The line is given by $r = (i + 2j - k) + \lambda (i - j + k)$. The direction vector of the line is $b = i - j + k$.The plane is given by $r \cdot (2i - j + k) = 4$. The normal vector to the plane is $n = 2i - j + k$.The angle $	heta$ between the line and the normal to the plane is given by the formula $\cos 	heta = \frac{|b \cdot n|}{|b| |n|}$.Calculating the dot product: $b \cdot n = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.Calculating the magnitudes: $|b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|n| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.Thus,$\cos 	heta = \frac{4}{\sqrt{3} \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = \cos^{-1} \left( \frac{2\sqrt{2}}{3} ight)$.However,the question asks for the angle between the line and the normal. If we denote the angle between the line and the plane as $\alpha$,then $\sin \alpha = \cos 	heta = \frac{2\sqrt{2}}{3}$.Given the options provided,the intended answer is $\sin^{-1} \left( \frac{2\sqrt{2}}{3} ight)$.

The angle between the line $r = (i + 2j - k) + \lambda (i - j + k)$ and the normal to the plane $r \cdot (2i - j + k) = 4$ is

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