Let →a and →b be two vectors such that |→a|=3, |→b|=6 and |→a+→b|=7. Then the value of (3→a−2→b).(2→a+5→b−4→a×→b) is
A
−306
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B
−295
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C
−284
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D
−290
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Solution
The correct option is C−284 (3→a−2→b)⋅(2→a+5→b−4→a×→b) =6|→a|2+15→a⋅→b−3→a⋅(4→a×→b)−4→a⋅→b−10|→b|2+2→b⋅(4→a×→b)
Now 3→a⋅(4→a×→b)=0 and 2→b⋅(4→a×→b)=0 because →a and →b will be perpendicular to 4→a×→b
So the equation reduces to (3→a−2→b)⋅(2→a+5→b−4→a×→b)=6|→a|2+15→a⋅→b−4→a⋅→b−10|→b|2 =6(3)2+11→a⋅→b−10(6)2 =11→a⋅→b−306⋯(1)
Now |→a+→b|=7 ⇒|→a+→b|2=72 ⇒|→a|2+2→a⋅→b+|→b|2=72 ⇒32+2→a⋅→b+62=72 ⇒→a⋅→b=2 Putting this value in equation (1), we get (3→a−2→b)⋅(2→a+5→b−4→a×→b)=11×2−306=−284