If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|$. Then the value of $|\vec{A}-\vec{B}|$ will be:

(A) $\sqrt{A^{2}+B^{2}-\sqrt{2} A B}$

(D) $\sqrt{A^{2}+B^{2}+\sqrt{2} A B}$

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Vector

A

and

B

are two vectors satisfying the relation

A \cdot B = ∣ A \times B ∣

. Then the value of

∣ A - B ∣

will be:

Answer: (A)

With

θ

the angle between the vectors,

A \cdot B = A B cos θ

and

∣ A \times B ∣ = A B sin θ

, so the given relation forces

tan θ = 1

, that is

θ = 4 5^{\circ}

. Then

∣ A - B ∣^{2} = A^{2} + B^{2} - 2 A B cos 4 5^{\circ} = A^{2} + B^{2} - 2 A B

∣ A - B ∣ = A^{2} + B^{2} - 2 A B

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option