To find the direction cosines we will use the formula \(cos\theta=\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\).
Let the vector \(u=\left<-2,0,3 \right>\).
So, the angle between the x-axis and the vector \(u\) is \(cos\alpha =\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\) and \(v=\left<1,0,0 \right>\).
Therefore,
\(cos\alpha =\frac{\left<-2,0,3 \right>\cdot \left<1,0,0 \right>}{\sqrt{-2^{2}+0^{2}+3^{2}}\sqrt{1^{2}+0+0}}\)
\(cos\alpha =\frac{-2(1)+0(0)+3(0)}{\sqrt{4+0+9}\sqrt{1}}\)
\(cos\alpha =\frac{-2+0+0}{\sqrt{13}\sqrt{1}}\)
\(cos\alpha =\frac{-2}{\sqrt{13}}\)
Now, the angle between the y-axis and the vector \(u\) is \(cos\beta =\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\) and \(v=\left<0,1,0 \right>\).
Therefore,
\(cos\beta =\frac{\left<-2,0,3 \right>\cdot \left<0,1,0 \right>}{\sqrt{-2^{2}+0^{2}+3^{2}}\sqrt{0+1^{2}+0}}\)
\(cos\beta =\frac{-2(0)+0(1)+3(0)}{\sqrt{4+0+9}\sqrt{1}}\)
\(cos\beta =\frac{0}{\sqrt{13}\sqrt{1}}\)
\(cos\beta =0\)
Now, the angle between the z-axis and the vector \(u\) is \(cos\gamma =\frac{u\cdot v}{\left\| u\right\|\left\|v \right\|}\) and \(v=\left<0,0,1 \right>\).
Therefore,
\(cos\gamma=\frac{\left<-2,0,3 \right>\cdot \left<0,0,1 \right>}{\sqrt{-2^{2}+0^{2}+3^{2}}\sqrt{0+0+1^{2}}}\)
\(cos\gamma=\frac{-2(0)+0(0)+3(1)}{\sqrt{4+0+9}\sqrt{1}}\)
\(cos\gamma=\frac{0+0+3}{\sqrt{13}\sqrt{1}}\)
\(cos\gamma=\frac{3}{\sqrt{13}}\)
