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B
→A×→B|→B|
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C
→A⋅→B|→A|
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D
→A⋅→B→B
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Solution
The correct option is A→A⋅→B|→B| Given, →A and →B
We know that the projection of →A along →B is given by →A.^B, where ^B is the unit vector of →B.
So, we have ^B=→B|→B|
Hence, projection is given by →A.^B=→A⋅→B|→B|