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Find the sine of the angle between the vectors 4i- 2j- 3k and 2i-3j+4k

Pranav , 8 Years ago
Grade 12th pass
anser 3 Answers
Arun
If we take dot product between these two vectors-
8 +6- 12 = |(4i-2j-3k)||(2i-3j+4k)| cos\theta
2/29 = cos\theta
So sin\theta=  \sqrt(837)/ 29
Last Activity: 8 Years ago
Shailendra Kumar Sharma
Dot product is best way to find out angle so 
Cos(theta) =A.B/(AB)
cos(theta) =(4j-2j-3k).(2i-3j+4k)/{sqrt(29)* sqrt(29)}
Cos(theta) =2/29
So sin(theta) =sqrt{1-cos2(theta)}
 =Sqer(837) /29
Last Activity: 8 Years ago
Namita kumari
-0.9
Last Activity: 8 Years ago
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