Question
Question: If\[\left| {A \times B} \right| = \sqrt 3 A \cdot B\], then the value of \[\left| {A \times B} \righ...
If∣A×B∣=3A⋅B, then the value of ∣A×B∣is:
A. (A2+B2+AB)21
B. (A2+B2+3AB)21
C. A +B
D. (A2+B2+3AB)21
Solution
Hint:
a. You should know vector calculus.
b. You should know vector identities.
c. You should know cosine values for 0,30, 45, 60, and 90.
Complete step by step solution:
We know that, by vector product,
(A×B)=ABsinθn^ - - - - - - - - - - - - - - - - - - - - (1)
And by Scalar product,
(A⋅B)=ABcosθ- - - - - - - - - - - - - - - - - - - - - - (2)
In question, the given quantities are
∣A×B∣=3A⋅B- - - - - - - - - - - - - - - - - - - - - - -(3)
By using
∣A×B∣=∣A∣∣B∣sinθ - - - - - - - - - - - - - - - - - - - - - (4)
We get,
∣A×B∣=ABsinθ - - - - - - - - - - - - - - - - - - - - - - - (5)
Now,
∣A⋅B∣=∣A∣∣B∣cosθ
A⋅B=ABcosθ - - - - - - - - - - - - - - - - - - - - - - - - - (6)
Substitute equation (6) in the right side of the equation (3) and equation (5) in the left side of the equation (3).
So,
ABsinθ=3ABcosθ
cosθsinθ=3
\tan \theta $$$$ = 3
⇒θ=60∘
Now,
(A+B)2=A2+B2+2A.B
(A+B)2=A2+B2+2ABcosθ
(A+B)2=A2+B2+2AB⋅21
(A+B)2=A2+B2+AB
Or,
A+B=(A2+B2+AB)21
Hence, Option (A) is correct
Note:
a. Substitution should not be reversed. If it reverses, all calculation will be wasted and time too.
b. Should be aware of the values of cos, sin and tan for 0, 30, 45, 60, and 90.
c. After getting the final calculation check the options and conveniently rearrange the solution.