Question

Find the angle between $\overrightarrow A $ and $\overrightarrow B $ if $\overrightarrow A + \overrightarrow B = \overrightarrow C $ and \[{A^2} + {B^2} = {C^2}\] ?

Answer 1

Hint: We know that the magnitude of addition of two vector is given by formula $\sqrt {{x^2} + {y^2} + 2xy\cos \theta } $ , where $\theta $ is the angle between the vector. Then, we will substitute the value of ${C^2}$ and cancel out the a and b terms. Them try to find the angle between $\overrightarrow A $ and $\overrightarrow B $.

Complete step by step answer:
We know that the vector addition is done on the triangle-law. When two vectors are represented by two sides of a triangle with the same order of magnitude and direction.We know that the magnitude of addition of two vector x and y is given by formula,
$\sqrt {{x^2} + {y^2} + 2xy\cos \theta } $
where $\theta $ is the angle between the vector x and y.

We have given $\overrightarrow A + \overrightarrow B = \overrightarrow C $,
So, we can write the magnitude of the vector is
$\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow C } \right|$
$\sqrt {{A^2} + {B^2} + 2AB\cos \theta } = C$
We will square both sides, then
${A^2} + {B^2} + 2AB\cos \theta = {C^2}$

We have given \[{A^2} + {B^2} = {C^2}\] , we will substitute ${C^2}$ in above equation and we get
${A^2} + {B^2} + 2AB\cos \theta = {A^2} + {B^2}$
$\cos \theta = 0$
As, we know that $\cos {90^ \circ } = 0$,
$\therefore \cos {90^ \circ } = 0$

Therefore, the angle between the vector $\overrightarrow A $ and $\overrightarrow B $ is $\theta = {90^ \circ }$.

Note: The magnitude and direction of a vector quantity are both the same. Vector subtraction and multiplication are also achievable using the triangle law of addition.Commutative and associative laws must be followed when adding vectors. We can't utilize a triangle-law of vector addition if two vectors are parallel or have an angle of 180 degrees.