If acute angle between the line r→=i→+2j→+l(4i→−3k→) and xy– plane is α and acute angle between theplanes x+2y=0,2x+y=0 is β then cos2⁡α+sin2⁡β=

sin⁡α=(4i¯−3k¯)⋅k¯5=35⇒cos⁡α=45⇒ Since ⇒cos⁡β=2+25⋅5⇒cos⁡β=45⇒sin⁡β=35

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If acute angle between the line $\vec{r} = \vec{i} + 2 \vec{j} + l (4 \vec{i} - 3 \vec{k})$ and xy– plane is $α$ and acute angle between the
planes $x + 2 y = 0, 2 x + y = 0$ is $β$ then $\cos^{2} α + \sin^{2} β =$

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$\sin α = \frac{(4 \bar{i} - 3 \bar{k}) \cdot \bar{k}}{5} = \frac{3}{5} \Rightarrow \cos α = \frac{4}{5}$

$\Rightarrow Since \Rightarrow \cos β = \frac{2 + 2}{\sqrt{5} \cdot \sqrt{5}}$

$\Rightarrow \cos β = \frac{4}{5} \Rightarrow \sin β = \frac{3}{5}$

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