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Question

Determine cosθ\cos \theta where θ\theta is the angle between u and v. u=2i3j+k, v=i2j+3ku = 2i - 3j + k, \ v = i - 2j + 3k

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We have to find cosθ\cos \theta where θ\theta is the angle between uu and vv and

u=2i3j+k,v=i2j+3k\begin{align*} u&=2i-3j+k, v=i-2j+3k \end{align*}

The vectors i,ji,j and kk are defined by

i=[100],j=[010],k=[00k]\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

u=2i3j+k=2[100]3[010]+[00k]=[231]v=i2j+3k=[100]2[010]+3[00k]=[123]\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 Theorem 7\textbf{\color{#c34632}Theorem 7}, θ\theta is the angle between uu and vv

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

The length is defined

u=(u12)+(u22)+(u32)=22+(3)2+12=4+9+1=14v=12+(2)2+13=1+4+9=14\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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