Question

Find the angle between A=2i+2j-k and B=6i-3j+2k

Answer 1

Let we have two planes given by vectors $A = A_1i+B_1j+C_1k$ and $B = A_2i+B_2j+C_2k$.

In our case

$A_1 = 2; B_1 = 2; C_1 = -1;$

$A_2 = 6; B_2 = -3; C_2 = 2$

The angle between the planes is equal to the angle between their normal vectors, it is enough to find the angle between the vectors $n_1(A_1,B_1,C_1)$ and $n_2(A_2,B_2,C_2)$.

That is

$\cos(\phi) = \cfrac{A_1A_2 + B_1B_2+C_1C_2 }{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}$

$\cos(\phi) = \cfrac{2*6 + 2*(-3)+(-1)*2 }{\sqrt{2^2+2^2+(-1)^2}\sqrt{6^2+(-3)^2+2^2}} = \cfrac{4}{\sqrt{9}\sqrt{49}} = \cfrac{4}{21} \approx 0.19047619$

$\phi = \arccos(0.19047619) \approx 79.0194 \degree$

Answer: the angle between $A=2i+2j-k$ and $B=6i-3j+2k$ is $79.0194 \degree$.