Find the angle between force $F = (3 \hat{i} + 4 \hat{j} - 5 \hat{k})$ units and displacement $d = (5 \hat{i} + 4 \hat{j} + 3 \hat{k})$ units. Also,find the projection of $F$ on $d$.

(N/A) The dot product of $F$ and $d$ is given by $F \cdot d = F_x d_x + F_y d_y + F_z d_z$.
$F \cdot d = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16$ units.
The magnitudes are $F = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ units and $d = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50}$ units.
Using $F \cdot d = F d \cos \theta$,we have $16 = \sqrt{50} \cdot \sqrt{50} \cdot \cos \theta$.
$16 = 50 \cos \theta \implies \cos \theta = \frac{16}{50} = 0.32$.
Therefore,$\theta = \cos^{-1}(0.32)$.
The projection of $F$ on $d$ is given by $\frac{F \cdot d}{|d|} = \frac{16}{\sqrt{50}} = \frac{16}{5\sqrt{2}} = \frac{16\sqrt{2}}{10} = 1.6\sqrt{2}$ units.