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

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(C) The equation of the line is $r = a + \lambda b$,where $b = -i + j + k$. The equation of the plane is $r \cdot n = d$,where $n = 3i + 2j - k$. The

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$\sin^{-1}\left(\frac{2}{\sqrt{42}}\right)$

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(C) The equation of the line is $r = a + \lambda b$,where $b = -i + j + k$. The equation of the plane is $r \cdot n = d$,where $n = 3i + 2j - k$. The angle $	heta$ between a line and a plane is given by $\sin 	heta = \frac{|b \cdot n|}{|b||n|}$. Here,$b \cdot n = (-1)(3) + (1)(2) + (1)(-1) = -3 + 2 - 1 = -2$. $|b| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$. $|n| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$. Thus,$\sin 	heta = \frac{|-2|}{\sqrt{3} \cdot \sqrt{14}} = \frac{2}{\sqrt{42}}$. Therefore,$	heta = \sin^{-1}\left(\frac{2}{\sqrt{42}}ight)$. Note: Since the options provided include $\sin^{-1}\left(\frac{-2}{\sqrt{42}}ight)$,we select the option that matches the magnitude calculation.

The angle between the line $r = (2i - j + k) + \lambda (-i + j + k)$ and the plane $r \cdot (3i + 2j - k) = 4$ is:

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