NIMCET 2015 — Mathematics PYQ
NIMCET | Mathematics | 2015If a and b are vectors in space, given by a=5i^−2j^ and b=142i^+j^+3k^, then the value of (2a+b)⋅[(a×b)×(a−2b)] is:
Choose the correct answer:
- A.
3
- B.
4
- C.
5
(Correct Answer) - D.
6
5
Explanation
Let E=(2a+b)⋅[(a×b)×(a−2b)]
Using property: u⋅(v×w)=[uvw]
E=[(2a+b)(a×b)(a−2b)]
Cyclic rearrangement:
E=(a×b)⋅[(a−2b)×(2a+b)]
Solving the cross product:
(a−2b)×(2a+b)=a×b−2b×2a
=a×b+4(a×b)=5(a×b)
Substituting back:
E=(a×b)⋅5(a×b)
E=5∣a×b∣2
Magnitude calculations:
∣a∣2=51+4=1
∣b∣2=144+1+9=1
a⋅b=70(1)(2)+(−2)(1)+0=0⟹a⊥b
Since θ=90∘:
∣a×b∣=∣a∣∣b∣sin90∘=1
Final Result:
E=5(1)2=5
Correct Option: (c)
Explanation
Let E=(2a+b)⋅[(a×b)×(a−2b)]
Using property: u⋅(v×w)=[uvw]
E=[(2a+b)(a×b)(a−2b)]
Cyclic rearrangement:
E=(a×b)⋅[(a−2b)×(2a+b)]
Solving the cross product:
(a−2b)×(2a+b)=a×b−2b×2a
=a×b+4(a×b)=5(a×b)
Substituting back:
E=(a×b)⋅5(a×b)
E=5∣a×b∣2
Magnitude calculations:
∣a∣2=51+4=1
∣b∣2=144+1+9=1
a⋅b=70(1)(2)+(−2)(1)+0=0⟹a⊥b
Since θ=90∘:
∣a×b∣=∣a∣∣b∣sin90∘=1
Final Result:
E=5(1)2=5
Correct Option: (c)