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JEE PYQ: Vector Algebra - Question ID 591535af6cd8 (JEE Main 2021)

ID: 591535af6cd8JEE Main 2021Single Correct MCQ
If A→\overrightarrow A and B→\overrightarrow B are two vectors satisfying the relation A→\overrightarrow A . B→\overrightarrow B = ∣A→×B→∣\left| {\overrightarrow A \times \overrightarrow B } \right|. Then the value of ∣A→−B→∣\left| {\overrightarrow A - \overrightarrow B } \right| will be :

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Step-by-step Explanation

Given, A→\overrightarrow A . B→\overrightarrow B = ∣A→×B→∣\left| {\overrightarrow A \times \overrightarrow B } \right| ..... (i)

Also, we know that

A→\overrightarrow A . B→\overrightarrow B = ∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right| cosθ\theta .... (ii)

and A→×B→\overrightarrow A \times \overrightarrow B = ∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right| sinθ\theta ..... (iii)

From Eqs. (i), (ii) and (iii), we get

∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right| cosθ\theta = ∣A→∣∣B→∣\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right| sinθ\theta

⇒cos⁡θ=sin⁡θ⇒sin⁡θcos⁡θ=1\Rightarrow \cos \theta = \sin \theta \Rightarrow {{\sin \theta } \over {\cos \theta }} = 1

⇒tan⁡θ=1\Rightarrow \tan \theta = 1

⇒tan⁡θ=tan⁡45∘⇒θ=45∘\Rightarrow \tan \theta = \tan 45^\circ \Rightarrow \theta = 45^\circ

∴\therefore ∣A→−B→∣=A2+B2−2∣A→∣∣B→∣cos⁡θ\left| {\overrightarrow A - \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta }

=A2+B2−2∣A→∣∣B→∣cos⁡(45∘)= \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos (45^\circ )}

=A2+B2−2AB= \sqrt {{A^2} + {B^2} - \sqrt 2 AB}
View Related Chapter: Vector AlgebraPractice Chapter Test (CBT Mode)

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