Question

Question #41db2

Answer 1

`36.7^0(1dp)`

Explanation:

we will use the dot (scalar) product of two vectors

`veca.vecb=|veca||vecb|costheta---(1)`

`theta=`the angle between the two vectors

also if

`veca=a_1hati+a_2hatj+a_3hatk`

`vecb=b_1hati+b_2hatj+b_3hatk`

`veca*vecb=a_1b_1+a_2b_2+a_3b_3---(2)`

we need to find the angle between

`veca=3hati+hatj-2hatk`

and +ve direction of the x-axis that is the vector

`vecb=hati`

`|veca|=sqrt(3^2+1^2+2^2)=sqrt14`

`|vecb|=1`

combing `(1)" & " (2)`

`(3hati+hatj-2hatk)*(hati)=sqrt14*1costheta`

`3xx1+0+0=sqrt14costheta`

`costheta=3/sqrt14`

`theta=cos^(-1)(3/sqrt14)=36.7^0( 1dp)`