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Grade: 12th pass

                        

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

3 years ago

Answers : (3)

Arun
24738 Points
							
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
3 years ago
Shailendra Kumar Sharma
188 Points
							
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
3 years ago
Namita kumari
23 Points
							-0.9
						
2 years ago
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