Question

FIND THE ANGLE BETWEEN two vectors A and B
if,
vec.A = i^ + 2j^ - k^
and
B = -i^ + j^ -2k^

Answer 1

Hope it helps you!
plzzz mark as brainliest.

Answer 2

You can use Dot Product of two vectors here.

$\vec{A} = \vec{i}+2\vec{j}-\vec{k}\\ \vec{B} = -\vec{i}+\vec{j}-2\vec{k}$
Dot Product:

$\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos(x)$
Therefore,

$( \vec{i}+2\vec{j}-\vec{k}) \cdot( -\vec{i}+\vec{j}-2\vec{k}) = | \vec{i}+2\vec{j}-\vec{k}||-\vec{i}+\vec{j}-2\vec{k}|\cos(x)$
If you know the two things:
The dot product is commutative and distributive, and i.i=j.j=k.k=1 and i.j=j.k=k.i=0
The dot product is easy to calculate and I am leaving it to you. (Here the dot product equals 3)

$3 = \sqrt{ { 1 }^{2} + {2}^{2} + {1}^{2} } \times \sqrt{ {1}^{2} + {1}^{2} + {2}^{2} } \cos(x) \\ 3 = \sqrt{6} \times \sqrt{6} \cos(x) \\ 3 = 6 \cos(x) \\ \frac{1}{2} = \cos(x) \\ \boxed{x = 60°}$