1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
295
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
284
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
290
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos