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If vecA = 2 hati + hatj + hatk and vecB...

If ` vecA = 2 hati + hatj + hatk and vecB = hati + hatj + hatk` are two vectores, then the unit vector is

A

perpendicular to `vecA "is" (-hatj+hatk)/(sqrt(2))`

B

parallel to `vecA "is" (2hati+hatj+hatk)/(sqrt(6))`

C

perpendicular to `vecB "is " (-hatj+hatk)/(sqrt(2))`

D

parallel to `vecA"is " (hati+hatj+hatk)/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`vecAxxvecB=|{:(hati,hatj,hatk),(2,1,1),(1,1,1):}|`
`=hati(1-1)-hatj(2-1)+hatk(2-1)=-hatj+hatk`
The unit vector perpendicular to `vecA` and `vecB` is `((-hatj+hatk)/(sqrt(2)))`.
So choices (a) and (c ) are correct.
Any vector whose magnitude is k ( constant ) times
`(2hati+hatj+hatk)` is parallel to `vecA` .
So unit vector `(2hati+hatj+hatk)/(sqrt(6))` is parallel to `vecA` .
So choice (b) is also correct.
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