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Question

Let ^u=u1^i+u2^j+u3^k be a unit vector in R3 and ^w=16(^i+^j+2^k). Given that there exists a vector v in R3 such that |^u×v|=1 and ^w(^u×v)=1. Which of the following statement(s) is(are) correct ?

A
There is exactly one choice for such v
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B
There are infinitely many choices for such v
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C
If ^u lies in the xy-plane then |u1|=|u2|
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D
If ^u lies in the xz-plane then 2|u1|=|u3|
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Solution

The correct option is C If ^u lies in the xy-plane then |u1|=|u2|
^u=u1^i+u2^j+u3^k
^w=16(^i+^j+2^k)
|^u×v|=1

^w(^u×v)=1
|^w| |^u×v|cosα=1cosα=1α=0
So, ^w^u and ^wv (i)
^u and v lies in the same plane.
So there will be infinite number of such v.

Option (3):
If ^u lies in the xy-plane,
^u=u1^i+u2^j^u^w=0 [From (i)]u1+u2=0|u1|=|u2|

Option (4):
If ^u lies in the xz-plane,
^u=u1^i+u3^k
^u^w=0u1+2u3=02|u3|=|u1|

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