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Question

Let a be a unit vector, b=2i+jk and c=i+3k. Then, maximum value of [abc] is

A
1
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B
10+6
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C
106
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D
59
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Solution

The correct option is D 59
Given, b=2i+jk and c=i+3k
Now, b×c=∣ ∣ ∣ijk211103∣ ∣ ∣
=i(30)j(6+1)+k(01)
=3i7jk
Now, abc=ab×c
=a(b×c)cosθ
=132+72+12cosθ
=59cosθ
abcmax=591
( maximum value of cosθ is 1)
Hence, maximum value is 59.

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