The position vectors of points A, B and C with respect to the origin are (8i - 2j), (2i + 6j) and (-10i + 4j) respectively. If ABCN is a parallelogram, find...

Answer Details

Position vectors: \(A=(8,-2),\ B=(2,6),\ C=(-10,4)\).

(a) In parallelogram \(ABCN\) the diagonals \(AC\) and \(BN\) bisect each other, so they share a midpoint.

Midpoint of \(AC=\left(\tfrac{8-10}{2},\tfrac{-2+4}{2}\right)=(-1,1)\).

Set midpoint of \(BN=(-1,1)\): \(\tfrac{2+N_x}{2}=-1\Rightarrow N_x=-4\); \(\tfrac{6+N_y}{2}=1\Rightarrow N_y=-4\).

\[N=-4\mathbf{i}-4\mathbf{j}\]

(b) \(\overrightarrow{AN}=N-A=(-4-8,\,-4+2)=(-12,-2)\); \(\overrightarrow{AB}=B-A=(2-8,\,6+2)=(-6,8)\).

(c) \(\overrightarrow{AN}\cdot\overrightarrow{AB}=(-12)(-6)+(-2)(8)=72-16=56\).

\(|\overrightarrow{AN}|=\sqrt{144+4}=\sqrt{148}\approx12.17\); \(|\overrightarrow{AB}|=\sqrt{36+64}=10\).

\[\cos\theta=\frac{56}{10\sqrt{148}}=0.4603\Rightarrow \theta\approx62.59^{\circ}\]