Find the angle between the two planes 2x + y | vedclass

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Question

(A) The angle $	heta$ between two planes is the angle between their normal vectors $\vec{N_1}$ and $\vec{N_2}$.Given the equations of the planes $2x

Vedclass · Accepted Answer

$	heta = \cos^{-1}\left(\frac{4}{21}ight)$

Vedclass · Answer

(A) The angle $	heta$ between two planes is the angle between their normal vectors $\vec{N_1}$ and $\vec{N_2}$.Given the equations of the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,the normal vectors are:$\vec{N_1} = 2\hat{i} + \hat{j} - 2\hat{k}$$\vec{N_2} = 3\hat{i} - 6\hat{j} - 2\hat{k}$The formula for the angle between two planes is $\cos 	heta = \left| \frac{\vec{N_1} \cdot \vec{N_2}}{|\vec{N_1}| |\vec{N_2}|} ight|$.First,calculate the dot product: $\vec{N_1} \cdot \vec{N_2} = (2)(3) + (1)(-6) + (-2)(-2) = 6 - 6 + 4 = 4$.Next,calculate the magnitudes:$|\vec{N_1}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.$|\vec{N_2}| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.Therefore,$\cos 	heta = \left| \frac{4}{3 	imes 7} ight| = \frac{4}{21}$.Thus,$	heta = \cos^{-1}\left(\frac{4}{21}ight)$.

Find the angle between the two planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$ using the vector method.

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