Question

the angle between vector A= iCAP+jCAP and vector B= iCAP-jCAP​

Answer 1

ANGLE BETWEEN THE VECTORS IS 90°.

Given:

$1) \: \vec{A} = \hat{i} + \hat{j}$

$2) \: \vec{B} = \hat{i} - \hat{j}$

To find:

Angle between the vectors ?

Calculation:

The angle between the vectors can be calculated by the use of DOT/SCALAR PRODUCT:

$\vec{A}. \vec{B} = (\hat{i} + \hat{j}).( \hat{i} - \hat{j})$

$\implies | \vec{A}| \times | \vec{B} | \times \cos( \theta) = (\hat{i} + \hat{j}).( \hat{i} - \hat{j})$

$\implies | \vec{A}| \times | \vec{B} | \times \cos( \theta) = (1 \times 1) + \{1 \times( - 1) \}$

$\implies | \vec{A}| \times | \vec{B} | \times \cos( \theta) = 1 - 1$

$\implies | \vec{A}| \times | \vec{B} | \times \cos( \theta) = 0$

$\implies \cos( \theta) = 0$

$\implies \theta= {90}^{ \circ}$

So, angle between vector is 90°.

Answer 2

Answer:

Cos theta=A.B/Magnitude of A .B

Explanation:

Cos theta =1×1+1×(-1/2

cos theta =0

theta=90 degrees