Question

Find the angle between the vectors αÌ…=2i+2j−k and\eta=6i−3j+2k.

a.\\(75^0\\) approximately
b.\\(78^0\\) approximately
c.\\(70^0\\) approximately
d.\\(79^0\\) approximately

Answer 1

$a=2i+2j-k,\;\;b=6i-3j+2k.$

$cos\theta=\frac{ab}{|a||b|}=\frac{2*6+2*(-3)+(-1)*2}{\sqrt{2^2+2^2+1^2}\sqrt{6^2+3^2+2^2}}=\frac{4}{21}$.

$\theta=arccos(\frac{4}{21})\approx79^0 .$

Answer d.