The angle between vecA = hati + hatj and...

The angle between `vecA = hati + hatj and vecB = hati - hatj` is

`45^(@)`

`90^(@)`

`-45^(@)`

`180^(@)`

Text Solution

Verified by Experts

The correct Answer is:

Given : `vecA = hati + hatj, vecB = hati - hatj`
`therefore|vecA| = sqrt((1)^(2) + (1)^(2)) = sqrt(2) and |vecB| = sqrt((1)^(2) + (-1)^(2)) = sqrt(2)`
let `theta` be angle between the vectors `vecA and vecB`. Then according to definition of scalar product (or dot product)
`vecA *vecB = |vecA||vecB| costheta or cos theta = (vecA*vecB)/(|vecA||vecB|) = ((hati+hatj)*(hati-hatj))/((sqrt(2))(sqrt(2)))= 0`
`therefore theta = cos^(-1) (0) = 90^(@)`

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What is the angle between vecA = 5hati - 5hatj and vecB = 5hati-5hatj ?

`90^@`

`45^@`

`0^@`

`60^@`

If angle between veca = hati - 2hatj + 3hatk and vecb = 2hati + hatj +hatk is theta then the value of sin theta is

`(3)/(2sqrt7)`

`(-2)/(sqrt7)`

`(4)/(3sqrt7)`

`(5)/(2sqrt7)`

vecA and vecB and vectors expressed as vecA = 2hati + hatj and vecB = hati - hatj Unit vector perpendicular to vecA and vecB is:

`(hati - hatj + hatk)/(sqrt(3)) `

`(hati + hatj - hatk)/(sqrt(3)) `

`(hati + hatj + hatk)/(sqrt(3)) `

`hatk`