The angle between the line frac{x - 1}{2} = f | vedclass

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(C) The direction vector of the line is $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.The normal vector to the plane $x + y + 4 = 0$ is $\vec{n} = 1\hat{i

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The direction vector of the line is $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.The normal vector to the plane $x + y + 4 = 0$ is $\vec{n} = 1\hat{i} + 1\hat{j} + 0\hat{k}$.The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(1) + (1)(1) + (-2)(0) = 2 + 1 + 0 = 3$.Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ and $|\vec{n}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$.Substituting these values: $\sin 	heta = \frac{|3|}{3 	imes \sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = \arcsin\left(\frac{1}{\sqrt{2}}ight) = 45^o$.

The angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{-2}$ and the plane $x + y + 4 = 0$ is ......... $^o$.

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