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Question: If <img src="https://cdn.pureessence.tech/canvas_610.png?top_left_x=300&top_left_y=983&width=207&hei...

If , are non zero vectors and is perpendicular to then a non zero vector satisfying . = a, for some scalar a,× = is:

A

αa⃗+(a⃗+b⃗)∣a⃗∣2\frac { \alpha \vec { a } + ( \vec { a } + \vec { b } ) } { | \vec { a } | ^ { 2 } }

B

αa⃗+a⃗×b⃗∣b⃗∣2\frac { \alpha \vec { a } + \vec { a } \times \vec { b } } { | \vec { b } | ^ { 2 } }

C

αa⃗−(a⃗×b⃗)∣a⃗∣2\frac { \alpha \vec { a } - ( \vec { a } \times \vec { b } ) } { | \vec { a } | ^ { 2 } }

D
Answer

αa⃗−(a⃗×b⃗)∣a⃗∣2\frac { \alpha \vec { a } - ( \vec { a } \times \vec { b } ) } { | \vec { a } | ^ { 2 } }

Explanation

Solution

a→\overrightarrow{a}× =

Ž × (×) = × or (.) – (.)=×