Vectors: collinear/coplanar, linear combination, linear dependence/independence

(b) 2N

Using $R^2=P^2+Q^2+2PQ\cos\theta$: $10=(\sqrt2)^2+Q^2+2(\sqrt2)Q\cos45^\circ$ $10=2+Q^2+2Q$ $Q^2+2Q-8=0 \Rightarrow Q=2\text{ N}$

Sine law: In any triangle $ABC$, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R,$ where $a,b,c$ are the sides opposite to angles $A,B,C$ and $R$ is the circumradius.

Lami’s theorem: If three coplanar forces $P,Q,R$ acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces: $\frac{P}{\sin\alpha}=\frac{Q}{\sin\beta}=\frac{R}{\sin\gamma},$ where $\alpha,\beta,\gamma$ are the angles between $(Q,R)$, $(R,P)$ and $(P,Q)$ respectively.

Proof: Since the three forces are in equilibrium, their vector sum is zero. Hence, when placed head to tail, they form a closed triangle whose sides are proportional to $P,Q,R$.

In this triangle, the angle opposite to side $P$ is $\pi-\alpha$, opposite to $Q$ is $\pi-\beta$, and opposite to $R$ is $\pi-\gamma$. By the sine law, $\frac{P}{\sin(\pi-\alpha)}=\frac{Q}{\sin(\pi-\beta)}=\frac{R}{\sin(\pi-\gamma)}.$ Since $\sin(\pi-\theta)=\sin\theta$, $\frac{P}{\sin\alpha}=\frac{Q}{\sin\beta}=\frac{R}{\sin\gamma}.$ This proves Lami’s theorem.