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Calculate the direction cosines of a→+b→, given that a→=4i^+7j^-5k^ and b→=3i^+4j^+k^

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Batolisisidn

The direction cosines of a→+b→, are

  • cos ∝ = 0.5232
  • cos β = 0.8065
  • cos γ = 0.2933

Given data:

a = 4i^ + 7j^ - 5k^

b= 3i^ + 4j^ + k^

Calculate the direction cosines of a→+b→

a→+b→ = 7i + 11j - 4k

( a→+b→ ) / | a→+b→ | = ( 7i + 11j - 4k ) / [tex]\sqrt{7^2 + 11^2 + 4^2 }[/tex]

                                = ( 7i + 11j - 4k ) / 13.64

                                = 0.5232 i + 0.8065 j + 0.2933 k

Therefore the direction cosines are

cos ∝ = 0.5232

cos β = 0.8065

cos γ = 0.2933

Hence we can conclude that the direction cosines of a→+b→ are as listed above.

Learn more about direction cosines : https://brainly.com/question/20734777

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