The angle between the planes 3x - 4y + 5z = 0 | vedclass

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(B) The angle $	heta$ between two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by $\cos 	heta = \left| \frac{a_

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$\frac{\pi}{2}$

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(B) The angle $	heta$ between two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by $\cos 	heta = \left| \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$.Here,the normal vectors are $\vec{n_1} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{n_2} = 2\hat{i} - 1\hat{j} - 2\hat{k}$.Substituting the values,we get $\cos 	heta = \left| \frac{(3)(2) + (-4)(-1) + (5)(-2)}{\sqrt{3^2 + (-4)^2 + 5^2} \sqrt{2^2 + (-1)^2 + (-2)^2}} ight|$.$\cos 	heta = \left| \frac{6 + 4 - 10}{\sqrt{9 + 16 + 25} \sqrt{4 + 1 + 4}} ight| = \left| \frac{0}{\sqrt{50} \sqrt{9}} ight| = 0$.Since $\cos 	heta = 0$,we have $	heta = \cos^{-1}(0) = \frac{\pi}{2}$.

The angle between the planes $3x - 4y + 5z = 0$ and $2x - y - 2z = 5$ is

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