Question

mth141 midterm test october 16, 2023
part ii full answer
give full solutions to each of questions 6, 7, 8, 9. only one solution for each
question is allowed.

1. (a) (4 marks) let \\( \vec{v}=(k, 1) \\) and \\( \vec{q}=(1,2) \\). find all the values of \\( k \\) such that the angle

between \\( \vec{v} \\) and \\( \vec{q} \\) is \\( \pi / 4 \\).
(b) (3 marks) let \\( \vec{v}=(k, 1) \\) and \\( \vec{q}=(1,2) \\). find \\( \operatorname{proj}_{\vec{q}} \vec{v} \\).

Explanation:

Step1: Use the formula for the cosine of the angle between two vectors

The formula for the cosine of the angle \(\theta\) between two vectors \(\vec{u}=(u_1,u_2)\) and \(\vec{v}=(v_1,v_2)\) is \(\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\vert\vec{u}\vert\vert\vec{v}\vert}\).
Given \(\vec{u}=(k,1)\), \(\vec{v}=(1,2)\), \(\vec{u}\cdot\vec{v}=k\times1 + 1\times2=k + 2\), \(\vert\vec{u}\vert=\sqrt{k^{2}+1}\), \(\vert\vec{v}\vert=\sqrt{1^{2}+2^{2}}=\sqrt{5}\), and \(\theta=\frac{\pi}{4}\) (so \(\cos\theta=\frac{\sqrt{2}}{2}\)).
We have \(\frac{k + 2}{\sqrt{k^{2}+1}\sqrt{5}}=\frac{\sqrt{2}}{2}\).

Step2: Square both sides of the equation

Squaring both sides gives \(\frac{(k + 2)^{2}}{5(k^{2}+1)}=\frac{2}{4}=\frac{1}{2}\).
Cross - multiply: \(2(k^{2}+4k + 4)=5(k^{2}+1)\).
Expand: \(2k^{2}+8k + 8 = 5k^{2}+5\).
Rearrange: \(3k^{2}-8k - 3=0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 3\), \(b=-8\), \(c=-3\)), the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) gives \(k=\frac{8\pm\sqrt{64+36}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}\).
So \(k=\frac{8 + 10}{6}=3\) or \(k=\frac{8-10}{6}=-\frac{1}{3}\).

Answer:

\(k = 3\) or \(k=-\frac{1}{3}\)