Question

Find the angle between the planes r.(2i - j + 2 k) = 3 and r.(3i + 6j + K) = 4.

Answer 1

Solution:

Angle between two planes given by

$\vec r.\vec n_{1}= d_{1}\\\\\vec r.\vec n_{2}= d_{2}$

$cos \: \theta = \frac{\vec n_{1}.\vec n_{2}}{ |\vec n_{1}| |\vec n_{2}| } \\ \\ cos \: \theta = \frac{2(3) + ( - 1)(6) + (2)(1)}{ \sqrt{ {2}^{2} + {1}^{2} + {2}^{2} } \sqrt{ {3}^{2} + {6}^{2} + {1}^{2} } } \\ \\ cos \: \theta = \frac{6 - 6 + 2}{ \sqrt{4 + 4 + 1} \sqrt{9 + 36 + 1} } \\ \\ = \frac{2}{3. \sqrt{46} } \\ \\ cos \: \theta = \frac{2}{3 \sqrt{46} } \\ \\ \theta = {cos}^{ - 1} ( \frac{2}{3 \sqrt{46} } )$

Answer 2

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