Question

Showing your work clearly, what is the angle between two vectors A and B if

A=3.0i - 4.0j
B=-2.0i+3.0k​

Answer 1

`\large\boxed{\large\boxed{70.6\º}}`

Step-by-step explanation:

1. Calculate the scalar product using the coordinates

`A\cdot B=(3.0\hat i-4.0\hat j+0\hat k)\cdot (-2.0\hat i+0\hat j+3.0\hat k)\\\\A\cdot B=(3)(-2)+(-4)(0)+(0)(3)=-6`

2. Write the scalar product using the cosine of the angle

`A\cdot B=|A|\cdot |B|\cdot cos\theta`

`|A|=√((3.0)^2+(-4.0)^2)=√(9.0+16.)=5.0`

`|B|=√((-2.0)^2+(3.0)^2)=√(4.0+9.0)=√(13.0)`

3. Equal the two scalar products and solve for cos(θ)

`-6.0=(5.0)(√(13.0))cos\theta\\\\cos\theta=0.3328\\\\\theta = arccos(0.3328)=70.6\º`