Find angle between A vector = 3icap + 4jcap and B vector = 12 j cap + 5 j cap
Answer (1)
9th Sep, 2020
Hello Aspirant
To find angle between two given vectors, we always use the dot product.
The dot product of two vectors is written as:
vector a.vector b = (mod of vector a)*(mod of vector b)*cos(theeta)
Now, a = 3icap + 4jcap, b = 12icap + 5jcap
So, we get vector a. vector b = (3icap + 4jcap)*(12icap + 5jcap) = 36 + 20 = 56
Also, mod of a = square root(9 + 16) = 5
mod of b = square root(144 + 25) = 13
So, we get 56 = 13*5*cos(theeta)
cos(theeta) = 56/65
theeta = cos - inverse(56/65)
So, the angle between given vectors is cos - inverse(56/65).
Comments (0)
Related Questions
the poyntings vector defines the?
67 Views
