## Related questions with answers

Determine $\cos \theta$ where $\theta$ is the angle between u and v. $u = 2i - 3j + k, \ v = i - 2j + 3k$

Solution

VerifiedWe have to find $\cos \theta$ where $\theta$ is the angle between $u$ and $v$ and

$\begin{align*} u&=2i-3j+k, v=i-2j+3k \end{align*}$

The vectors $i,j$ and $k$ are defined by

$\begin{align*} i&=\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, j=\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}, k=\begin{bmatrix} 0\\ 0\\ k \end{bmatrix} \end{align*}$

Then

$\begin{align*} u&=2i-3j+k\\ &=2\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}-3\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}+\begin{bmatrix} 0\\ 0\\ k \end{bmatrix}=\begin{bmatrix} 2\\ -3\\ 1 \end{bmatrix}\\ v&=i-2j+3k\\ &=\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}-2\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}+3\begin{bmatrix} 0\\ 0\\ k \end{bmatrix}=\begin{bmatrix} 1\\ -2\\ 3 \end{bmatrix} \end{align*}$

According to $\textbf{\color{#c34632}Theorem 7}$, $\theta$ is the angle between $u$ and $v$

$\begin{align*} u \cdot v=||u|| ||v|| \cos \theta \end{align*}$

The length is defined

$\begin{align*} ||u||&=\sqrt{(u^{2}_{1})+(u_{2}^{2})+(u_{3}^{2})}\\ &=\sqrt{2^{2}+(-3)^{2}+1^{2}}\\ &=\sqrt{4+9+1}\\ &=\sqrt{14} ||v||&=\sqrt{1^{2}+(-2)^{2}+1^{3}}\\ &=\sqrt{1+4+9}\\ &=\sqrt{14} \end{align*}$

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