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Let bar(a)= hat(i) + hat(j) + hat(k), ba...

Let `bar(a)= hat(i) + hat(j) + hat(k), bar(b)= hat(i) + 3hat(j) + 5hat(k) and bar(c )= 7hat(i) + 9hat(j) + 11hat(k)`. Then, the area of the parallelogram with diagonals `bar(a) + bar(b) and bar(b) + bar(c )` is

A

`4 sqrt6`

B

`(1)/(2) sqrt21`

C

`(sqrt6)/(2)`

D

`sqrt6`

Text Solution

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The correct Answer is:
A
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