Question

q5. the angle between the two vectors $vec{a_1}=\begin{bmatrix}1\\2\\3end{bmatrix}$ and $vec{a_2}=\begin{bmatrix}-2\\-4\\-6end{bmatrix}$ is equal to: a) 0 b) $pi$ c) $\frac{pi}{2}$ d) $\frac{3pi}{2}$

Explanation:

Step1: Recall dot - product formula

The dot - product formula is $\vec{a_1}\cdot\vec{a_2}=\vert\vec{a_1}\vert\vert\vec{a_2}\vert\cos\theta$, where $\theta$ is the angle between $\vec{a_1}$ and $\vec{a_2}$, $\vec{a_1}\cdot\vec{a_2}=a_{11}a_{21}+a_{12}a_{22}+a_{13}a_{23}$, $\vert\vec{a_1}\vert=\sqrt{a_{11}^2 + a_{12}^2+a_{13}^2}$, and $\vert\vec{a_2}\vert=\sqrt{a_{21}^2 + a_{22}^2+a_{23}^2}$.
First, calculate $\vec{a_1}\cdot\vec{a_2}$:
$\vec{a_1}\cdot\vec{a_2}=(1)\times(- 2)+(2)\times(-4)+(3)\times(-6)=-2-8 - 18=-28$.

Step2: Calculate magnitudes

$\vert\vec{a_1}\vert=\sqrt{1^2 + 2^2+3^2}=\sqrt{1 + 4+9}=\sqrt{14}$.
$\vert\vec{a_2}\vert=\sqrt{(-2)^2+(-4)^2+(-6)^2}=\sqrt{4 + 16+36}=\sqrt{56}=2\sqrt{14}$.

Step3: Find $\cos\theta$

Substitute into $\vec{a_1}\cdot\vec{a_2}=\vert\vec{a_1}\vert\vert\vec{a_2}\vert\cos\theta$:
$-28=\sqrt{14}\times2\sqrt{14}\cos\theta$.
$-28 = 2\times14\cos\theta$.
$-28=28\cos\theta$.
$\cos\theta=-1$.

Step4: Determine $\theta$

Since $\cos\theta=-1$ and $0\leq\theta\leq\pi$, then $\theta = \pi$.

Answer:

b) $\pi$