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Question

If α =3i^+4j^+5k^ and β =2i^+j^-4k^, then express β in the form of β =β1+β2, where β1 is parallel to α and β2 is perpendicular to α .

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Solution

Given that α=3i^+4j^+5k^ and β=2i^+j^-4k^ Also,β=β1+β2β2=β-β1 ... 1Since β1 is parallel to α,β1=t αβ1=t 3i^+4j^+5k^ =3t i^+4t j^+5t k^ ...(2)Substituting the values of β1 andα in (1), we getβ2=2i^+j^-4k^ -3t i^+4t j^+5t k^=2-3t i^+1-4tj^+-4-5t k^ ... 3Since β2 is perpendicular to α,β2. α=02-3t i^+1-4tj^+-4-5t k^. 3i^+4j^+5k^=03 2-3t +4 1-4t+5 -4-5t=06-9t+4-16t-20-25t=0-50t=10t=-15From (2) and (3), we getβ1=-15 3i^+4j^+5k^ β2=135i^+95j^-3k^=1513i^+9j^-15k^

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