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 \(u=\left<-1,1,-4 \right>\).
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,
\(cos\alpha =\frac{\left<-1,1,-4 \right>\cdot \left<1,0,0 \right>}{\sqrt{(-1)^{2}+1^{2}+(-4)^{2}}\sqrt{1^{2}+0+0}}\)
\(cos\alpha =\frac{-1(1)+1(0)-4(0)}{\sqrt{1+1+16}\sqrt{1}}\)
\(cos\alpha =\frac{-1+0+0}{\sqrt{18}\sqrt{1}}\)
\(cos\alpha =\frac{-1}{\sqrt{18}}\)
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,
\(cos\beta =\frac{\left<-1,1,-4 \right>\cdot \left<0,1,0 \right>}{\sqrt{(-1)^{2}+1^{2}+(-4)^{2}}\sqrt{0+1^{2}+0}}\)
\(cos\beta =\frac{-1(0)+1(1)-4(0)}{\sqrt{1+1+16}\sqrt{1}}\)
\(cos\beta =\frac{0+1+0}{\sqrt{18}\sqrt{1}}\)
\(cos\beta =\frac{1}{\sqrt{18}}\)
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,
\(cos\gamma=\frac{\left<-1,1,-4 \right>\cdot \left<0,0,1 \right>}{\sqrt{(-1)^{2}+1^{2}+(-4)^{2}}\sqrt{0+0+1^{2}}}\)
\(cos\gamma=\frac{-1(0)+1(0)-4(1)}{\sqrt{1+1+16}\sqrt{1}}\)
\(cos\gamma=\frac{0+0-4}{\sqrt{18}\sqrt{1}}\)
\(cos\gamma=\frac{-4}{\sqrt{18}}\)
