Question

Please answer this...This is very urgent....

Answer 1

Given : Vectors 2i + 2j + 4k , 3i + j + 2k

To find : Sinθ where θ is angle between vectors

Solution:

2i + 2j + 4k

3i + j + 2k

Sinθ = | A × B | / | A | | B|

A × B = = i(4 - 4) -j(4 - 12) + k (2 - 6) = 8j - 4k

| A × B | = √0² + 8² + 4² = √80

| A | = √2² + 2² + 4² = √24

|B | = √3² + 1² + 2² = √14

Sinθ = √80 / √24√14

=> Sinθ = √80 / √336

=> Sinθ = √5 / √21

or Cosθ = ( 2 * 3 + 2 * 1 + 4 * 2)/ | A | | B|

= 16/ √336

= 4/√21

Cos²θ = 16/21

=> 1 - Cos²θ = 1 - 16/21

=> Sin²θ = 5/21

=> Sinθ = √5 / √21

none of the given option matches

Seems mistake in data

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²

Answer 2

$copied \: from \: the \: answer \: above$