# What is the angle between the two vectors A and B, where A=2i+3j+4k and B=i-2j+3k

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### 2 Answers

To solve this problem we use the scalar or dot product of vectors. Now for any two vectors A and B which have an angle x between them, A dot B = |A|*|B|*cos x, where |A| and |B| denote the magnitude of the vectors A and B.

In terms of the x, y and z components of A and B, A dot B is given as Ax*Bx + Ay*By + Az*Bz.

So A dot B = Ax*Bx + Ay*By + Az*Bz = |A|*|B|*cos x

=> cos x = [Ax*Bx + Ay*By + Az*Bz] / [|A|*|B|] … (1)

Now the vectors we have are A= 2i + 3j + 4k and B= i – 2j + 3k.

So Ax = 2, Ay = 3 and Az = 4 and Bx = 1, By = -2 and Bz = 3

Substituting the values of the axial components in (1) we get

cos x = [2*1 + 3*(-2) + 4*3]/ [sqrt (2^2 + 3^2 + 4^2)* sqrt (1^2 + (-2) ^2 + 3^2)]

= (2 – 6 +12) / [sqrt (4 + 9 + 16)* sqrt (1 + 4 + 9)]

= 8 / (sqrt 29 * sqrt 14)

= 0.3970

As cos x = 0.3970

x = arccos (0.3970) = 66.6 degrees.

**Therefore the angle between the two vectors is 66.6 degrees.**

We know that the cross product of the vectors , A and B is given by: AXB = [(i j k),( 2 , 3 4) , (1 , -2 3)]

So AXB = (3*3- (-2)*4)i - (2*3-1*4)j + (2*(-2) - 1*3)k.

Therefore AXB = 17i +2j -7k.

Therefore |AXB| = |17i+2j-7k| = sqrt(17^2+2^2+(-7)^2} = sqrt(342)....(1)

|A| = |2i+3j+4k| = sqrt(2^2+3^2+4^2) = sqrt29... (2)

|B| = | i-2i+3k| = sqrt(1^2+(-2)^2 +3^2) = sqrt14 ..(3)

Also we know the relation ,|AXB| = |A|*|B| sinx.

Therefore sinx = |AXB|/(|A|*|B|)

sinx = sqrt 342/(sqrt29*sqrt14)

sinx = sqrt {342/(29*14)}

x = arc sin {342/(29*14)}^(1/2)

x= 66.61 deg approximately, or x = 0.37 radians (nearly