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Let vecu=2hati-hatj+hatk, vecv=-3hatj+2h...

Let `vecu=2hati-hatj+hatk, vecv=-3hatj+2hatk` be vectors and `vecw` be a unit vector in the xy-plane. Then the maximum possible value of `|(vecu xx vecv)|.|vecw|` is

A

`sqrt(5)`

B

`sqrt(12)`

C

`sqrt(13)`

D

`sqrt(17)`

Text Solution

Verified by Experts

The correct Answer is:
D
`vecu xx vecv = (2hati-hatj+hatk) xx (-3hatj+2hatk)`
`=hati-4hatj-6hatk`
Let `vecw = ahati+bhatj`
We have `a^(2)+b^(2)=1`
So let `a=costheta,b=sintheta`
Now, `vecu xx vecv. vecw=a-4b=costheta-4sintheta`
Max. value `=sqrt(1^(2)+(-4)^(2))=sqrt(17)`
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