Tip:A–D to answerE for explanationV for videoS to reveal answer
If a and b are vector in space , given by a=5i^−2j^ and b=142i^+j^+3k^, then the value of (2a+b)⋅[(a×b)×(a−2b)] is
- A.
3
- B.
4
- C.
5
(Correct Answer) - D.
6
Explanation
−(2a+b)⋅[(a−2b)×(a×b)]
=−(2a+b)⋅[(a−2b)⋅ba−(a−2b)⋅ab]
=(2a+b)⋅[(a−2b)⋅ab−(a−2b)⋅ba]
=(2a+b)⋅[∣a∣2−2b⋅ab−a⋅b−2∣b∣2a]
Now, a⋅b=(5i^−2j^)⋅(142i^+j^+3k^)=2−2=0
∣a∣=1,;∣b∣=1
=(2a+b)⋅[1−0b−0−2a]
=(2a+b)⋅[b+2a]
=(2a+b)⋅[b+2a]
=2a⋅b+4∣a∣2+∣b∣2+4a⋅b
=0+4+1+0=5
Explanation
−(2a+b)⋅[(a−2b)×(a×b)]
=−(2a+b)⋅[(a−2b)⋅ba−(a−2b)⋅ab]
=(2a+b)⋅[(a−2b)⋅ab−(a−2b)⋅ba]
=(2a+b)⋅[∣a∣2−2b⋅ab−a⋅b−2∣b∣2a]
Now, a⋅b=(5i^−2j^)⋅(142i^+j^+3k^)=2−2=0
∣a∣=1,;∣b∣=1
=(2a+b)⋅[1−0b−0−2a]
=(2a+b)⋅[b+2a]
=(2a+b)⋅[b+2a]
=2a⋅b+4∣a∣2+∣b∣2+4a⋅b
=0+4+1+0=5
