If vectors `vecA` and `vecB` are such that `|vecA_vecB|=|vecA|=|vecB|`, then `|vecA-vecB|` may be equated to

Options

A. `(sqrt3)/(2) |vecA|`
B. `|vecA|`
C. `sqrt2|vecA|`
D. `sqrt3|vecA|`

Answer

4

Explanation

If `|vecA+vecB|= |vecA| = |vecB|`
then angle between them `theta = 120^(@)`
Let `|vecA|= |vecB|=a`
So `|vecA-vecB| = sqrt(A^(2) + B^(2) - 2AB cos theta )`
`" "= sqrt(a^(2) + a^(2) - 2a^(2) (-1//2)) = sqrt3|vecA|`