Example

Question:

Find the angle between force
\(\vec{F} = (3\hat{\imath} + 4\hat{\jmath} - 5\hat{k})\) unit and displacement
\(\vec{d} = (5\hat{\imath} + 4\hat{\jmath} + 3\hat{k})\) unit.
Also find the projection of \(\vec{F}\) on \(\vec{d}\).

Solution:

\[ \vec{F} \cdot \vec{d} = F_x d_x + F_y d_y + F_z d_z = 3 \times 5 + 4 \times 4 + (-5) \times 3 = 16~\text{unit} \] \[ \vec{F} \cdot \vec{d} = F d \cos\theta = 16~\text{unit} \] Now, \[ \vec{F} \cdot \vec{F} = F^2 = F_x^2 + F_y^2 + F_z^2 = 9 + 16 + 25 = 50~\text{unit} \] \[ \vec{d} \cdot \vec{d} = d^2 = d_x^2 + d_y^2 + d_z^2 = 25 + 16 + 9 = 50~\text{unit} \] \[ \cos\theta = \frac{16}{\sqrt{50 \times 50}} = \frac{16}{50} = 0.32 \] \[ \theta = \cos^{-1}(0.32) \]
The projection of \(\vec{F}\) on \(\vec{d}\) is \[ |\vec{F}| \cos\theta = 7.07 \times 0.32 = 2.26~\text{unit} \quad (\text{where } |\vec{F}| = \sqrt{50} \approx 7.07) \]