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Physics Question on System of Particles & Rotational Motion

Three vectors satisfy the relation A→.B→=0\overrightarrow{A}.\overrightarrow{B}=0 and A→.C→=0\overrightarrow{A}.\overrightarrow{C}=0 then A→\overrightarrow{A} is parallel to

A

B→×C→\overrightarrow{B}\times\overrightarrow{C}

B

B→.C→\overrightarrow{B}.\overrightarrow{C}

C

C→\overrightarrow{C}

D

B→\overrightarrow{B}

Answer

B→×C→\overrightarrow{B}\times\overrightarrow{C}

Explanation

Solution

Here A→.B\overrightarrow{A} .{B} =AB cosθAB\, cos \theta = AB cos90∘AB\, cos 90^\circ = O ⇒\Rightarrow A→⊥B→\overrightarrow{A} \perp \overrightarrow{B} Similarly, A→⊥C→⇒B→\overrightarrow{A} \perp \overrightarrow{C} \Rightarrow \overrightarrow{B} and C→ \overrightarrow{C} are in the same plane and A→ \overrightarrow{A} is perpendicular to them. Thus A→∣∣B→×C→ \overrightarrow{A} || \overrightarrow{B} \times \overrightarrow{C}