The acute angle between the line r = (-hat{i} | vedclass

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(A) The line is given by $r = a + \lambda b$,where $b = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The plane is given by $r \cdot n = d$,where $n = 10\hat{i} + 2

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{8}{21}\right)$

Vedclass · Answer

(A) The line is given by $r = a + \lambda b$,where $b = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The plane is given by $r \cdot n = d$,where $n = 10\hat{i} + 2\hat{j} - 11\hat{k}$.The angle $	heta$ between a line with direction vector $b$ and a plane with normal vector $n$ is given by $\sin 	heta = \frac{|b \cdot n|}{|b||n|}$.First,calculate the dot product $b \cdot n = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40$.Next,calculate the magnitudes $|b| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.And $|n| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.Substituting these values into the formula:$\sin 	heta = \frac{|-40|}{7 	imes 15} = \frac{40}{105} = \frac{8}{21}$.Therefore,$	heta = \sin^{-1}\left(\frac{8}{21}ight)$.

The acute angle between the line $r = (-\hat{i} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and the plane $r \cdot (10\hat{i} + 2\hat{j} - 11\hat{k}) = 3$ is:

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