Question #f8c71

1 Answer
Oct 25, 2017

The angle is #=arccos(1/sqrt(k^2+2))#

Explanation:

Let #vecA= <1,k,1>#

#vecB=<1,0,0>#

The angle between #vecA# and #vecB# is given by the dot product definition.

#vecA.vecC=∥vecA∥*∥vecB∥costheta#

Where #theta# is the angle between #vecA# and #vecB#

The dot product is

#vecA.vecB=〈1,k,1〉.〈1,0,0〉=1+0+0=1#

The modulus of #vecA#= #∥〈1,k,1〉∥=sqrt(1+k^2+1)=sqrt(k^2+2)#

The modulus of #vecB#= #∥〈1,0,0〉∥=sqrt(1+0+0)=sqrt1#

So,

#costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=1/(sqrt(k^2+2)*sqrt1)=1/sqrt(k^2+2)#

#theta=arccos(1/sqrt(k^2+2))#