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What is the component of 3i^+4j^ along i^+j^

Asked by Sunita 11th May 2019, 11:51 AM
Answered by Expert
Answer:
component of vector (3i + 4j) along (i+j) is (3i+4j)cosθ, where θ is angle between the vectors (3i+4j) and (i+j)
 
we have begin mathsize 12px style cos theta space equals space fraction numerator open parentheses 3 i space plus space 4 j close parentheses times open parentheses i plus j close parentheses over denominator open vertical bar open parentheses 3 i space plus space 4 j close parentheses close vertical bar space open vertical bar open parentheses i plus j close parentheses close vertical bar end fraction space equals space fraction numerator 7 over denominator 5 square root of 2 end fraction almost equal to space 0.99 end style
hence component of vector (3i + 4j) along (i+j) is 0.99(3i+4j)
Answered by Expert 11th May 2019, 1:58 PM
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