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The sides of a parallelogram are 2hati +...

The sides of a parallelogram are `2hati +4hatj -5hatk and hati + 2hatj +3hatk `. The unit vector parallel to one of the diagonals is

A

`(1)/(7)(3hati +6hatj-2hatk)`

B

`(1)/(7)(3hati-6hatj -2hatk)`

C

`(1)/(sqrt(69))(hati +2hatj +8hatk)`

D

`(1)/(sqrt(69))(-hati -2hatj +8hatk)`

Text Solution

Verified by Experts

The correct Answer is:
A, D
Let `veca = 2hati + 4hatj - 5hatk and vecb = hati + 2hatj +3hatk`.
Then the diagonals of the parallelogram are
`vecp = veca + vecb and vecq = vecb - veca`,
i.e., `vecp = 3hati +6hatj-2hatk, vecq =-hati -2hatj +8hatk`
So, unit vectors along the diagonals are
`(1)/(7) ( 3hati + 6hatj - 2hatk) and (1)/(sqrt(69))(-hati-2hatj+8hatk)`.
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