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Let vecV=2hati+hatj-hatk and vecW=hati+3...

Let `vecV=2hati+hatj-hatk` and `vecW=hati+3hatk`. It `vecU` is a unit vector, then the maximum value of the scalar triple product `[(vecU, vecV, vecW)]` is

A

`-1`

B

`sqrt10 + sqrt6`

C

`sqrt59`

D

`sqrt60`

Text Solution

Verified by Experts

The correct Answer is:
c
Given that `vecV = 2hati +hatj -hatk and vecW =hati + 3hatk and vecU` is a unit vector
` |vecU|=1`
Now `|vecU vecV vecW] = vecU.(vecV xx vecW)`
`= vecU . (2hati +hatj -hatk) xx ( hati + 3hatk)`
`vecU . (3hati -7hatj - hatk)`
` sqrt(3^(2)+7^(2)+ 1^(2)) cos theta`
Which is maximum when `cos theta =1`
therefore, maximum value of [`vecU vecV vecW] - sqrt59`
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