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Question

# Let →a=^i+^j+√2^k,→b=b1^i+b2^j+√2^k and →c=5^i+^j+√2^k be three vectors such that the projection vector of →b on →a is →a. If →a+→b is perpendicular to →c, then |→b| is equal to :

A
6
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B
4
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C
22
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D
32
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Solution

## The correct option is A 6Given, Projection of →b on →a is →a. ∴→b.→a|→a|=|→a| ⇒b1+b2+2√1+1+2=√1+1+2 ⇒ b1+b2=2 ...(1) Since, →a+→b is perpendicular to →c So, (→a+→b)⋅→c=0 ⇒→a⋅→c+→b⋅→c=0 ⇒(5+1+2)+(5b1+b2+2)=0 ⇒5b1+b2=−10 ...(2) From Equations (1) and (2), b1=−3, b2=5 ∴ →b=−3^i+5^j+√2^k Hence, |→b|=√9+25+2 ⇒ |→b|=6

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