Question

>Find the component of vector i+3j-2k along the direction of vector i+j.​

Answer 1

$\sqrt{2 ^{2} } + \sqrt{ {4}^{2} } + \sqrt{ 2 }^{2} } = 2 \sqrt{6} \\ \ \\ 2i + 4j - 2k \\ \\ cos \alpha = \frac{2}{2 \sqrt{6} } \sqrt{5 } \: \\ \\ cos \beta = \frac{4}{2 \sqrt{6} } \\ \\ cos \gamma = \frac{ -2}{2 \sqrt{6} }$

Answer 2

Answer:

god

Explanation: