Chapter 5.5, Problem 1AP

For Problem 1-2, determine the angle between the given vectors $u$ and $v$ using the standard inner product on ${ℝ}^n$.

$u=\left(2,3\right)$ and $v=\left(4,-1\right)$

To determine

The angle between the given vectors $u$ and $v$.

Answer to Problem 1AP

Solution:

The angle between the vectors is $\theta ={\cos }^{-1}\left(\frac{5}{\sqrt{221}}\right)\approx 1.23\text{rad}$.

Explanation of Solution

Given:

The vectors are $u=\left(2,3\right)$ and $v=\left(4,-1\right)$

Approach:

The angle between two non zero vectors $u$ and $v$ is given by,

$\cos \theta =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$ …$\left(1\right)$

If $u=\left(a_1,a_2.a_3\right)$ and $v=\left(b_1,b_2,b_3\right)$ are vectors, then the dot product of $u$ and $v$ is denoted by $u\cdot v$, and defined by,

$u\cdot v=a_1{b}_1+a_2{b}_2+a_3{b}_3$ …$\left(2\right)$

The magnitude of the vector $u=\left(a_1,a_2.a_3\right)$ is given by,

$\Vert u\Vert =\sqrt{a_1^2+a_2^2+a_3^2}$ …$\left(3\right)$

Calculation:

Substitute $\left(2,3\right)$ for $u$ and $\left(4,-1\right)$ for $v$ in equation $\left(2\right)$

$\begin{array}{rll}u\cdot v& =& 2\left(4\right)+3\left(-1\right)\\ & =& 8-3\\ & =& 5\end{array}$

Substitute $\left(2,3\right)$ for $u$ in equation $\left(3\right)$

$\begin{array}{rll}\Vert u\Vert & =& \sqrt{2^2+3^2}\\ & =& \sqrt{4+9}\\ & =& \sqrt{13}\end{array}$

Substitute $\left(2,3\right)$ for $v$ in equation $\left(3\right)$.

$\begin{array}{rll}\Vert v\Vert & =& \sqrt{4^2+{\left(-1\right)}^2}\\ & =& \sqrt{16+1}\\ & =& \sqrt{17}\end{array}$

Substitute $5$ for $u\cdot v$, $\sqrt{13}$ for $\Vert u\Vert$ and $\sqrt{17}$ for $\Vert v\Vert$ in equation $\left(1\right)$.

$\begin{array}{rll}\cos \theta & =& \frac{5}{\sqrt{13}\sqrt{17}}\\ \theta & =& {\cos }^{-1}\frac{5}{\sqrt{221}}\end{array}$

Therefore, the angle between the vectors is $\theta ={\cos }^{-1}\left(\frac{5}{\sqrt{221}}\right)\approx 1.23\text{rad}$.

Conclusion:

Hence, the angle between the vectors is $\theta ={\cos }^{-1}\left(\frac{5}{\sqrt{221}}\right)\approx 1.23\text{rad}$.