A×B = √3 A.B, then the value of |A+B| is
Answers
Answered by
2
Explanation:
ANSWER
We know that,
A×B=ABsinθ
Also, A.B=ABcosθ
Given : ∣A×B∣=
3
A.B
Using ∣A×B∣=∣A∣∣B∣sinθ
We get ∣A×B∣=ABsinθ
A.B=∣A∣∣B∣cosθ
∴ ABsinθ=
3
ABcosθ
tanθ=
3
⟹θ=60
o
Now (A+B)
2
=A
2
+B
2
+2A.B
=A
2
+B
2
+2ABcosθ
=A
2
+B
2
+2AB×
2
1
=A
2
+B
2
+AB
or ∣A+B∣=(A
2
+B
2
+AB)
1/2
Answered by
1
Answer:
(A² + B² + AB)^1/2
Explanation:
We know,
X̄ × Ȳ = XY sinθ n̂
=> | X̄ × Ȳ | = XY sinθ
& | X̄ · Ȳ | = XY cosθ
Therefore,
AB sinθ = √3 AB cosθ (from qs)
=> sinθ = √3 cosθ
=> tanθ = √3
=> θ = arctan(√3) = π/3ᶜ
Again,
| X̄ + Ȳ | = √(X² + Y² + 2XYcosθ)
=> | A vec + B vec | = √(A² + B² + 2ABcos(π/3))
= √(A² + B² + 2AB(1/2)) = √(A² + B² + AB).
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