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53. Determine the sine of the angle between thevectors (3i + 2j +4k) and (2i -2j- 4k).

Answer»

Dot product of two vectors is (3i+2j+4k)•(2i-2j-4k) = (6-4-16) = -14

now the magnitude of these vectors are √(3)²+(2)²+(4)² = √9+4+16 = √29 and √(2)²+(-2)²+(-4)² = √24 = 2√6

now angle between them is

cos∅ = dot product/(magnitude1*magnitude2) = -14/(2√6*√29) = -7/√174 ∅ = cos-¹(-7/√174)



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