Q. 46 - Calculus

Question

Find the direction angles and direction cosines for the vectors given in Exercises 43–46.

\(\left<-3,4,2 \right>\)

Step-by-Step Solution

Answer

The direction angles and direction cosines for the vectors are

\(cos\alpha =\frac{-3}{\sqrt{29}}, \alpha =cos^{-1}\left ( \frac{-3}{\sqrt{29}} \right )\)

\(cos\beta =\frac{4}{\sqrt{29}}, \beta=cos^{-1}\left ( \frac{4}{\sqrt{29}} \right )\) 

\(cos\gamma =\frac{2}{\sqrt{29}}, \gamma=cos^{-1}\left ( \frac{2}{\sqrt{29}} \right )\) 

Step 1. Find the direction cosines

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 \(u=\left<-3,4,2 \right>\).

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,

\(cos\alpha =\frac{\left<-3,4,2 \right>\cdot \left<1,0,0 \right>}{\sqrt{(-3)^{2}+4^{2}+2^{2}}\sqrt{1^{2}+0+0}}\)

\(cos\alpha =\frac{-3(1)+4(0)+2(0)}{\sqrt{9+16+4}\sqrt{1}}\)

\(cos\alpha =\frac{-3+0+0}{\sqrt{29}\sqrt{1}}\)

\(cos\alpha =\frac{-3}{\sqrt{29}}\)

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,

\(cos\beta  =\frac{\left<-3,4,2 \right>\cdot \left<0,1,0 \right>}{\sqrt{(-3)^{2}+4^{2}+2^{2}}\sqrt{0+1^{2}+0}}\)

\(cos\beta  =\frac{-3(0)+4(1)+2(0)}{\sqrt{9+16+4}\sqrt{1}}\)

\(cos\beta  =\frac{0+4+0}{\sqrt{29}\sqrt{1}}\)

\(cos\beta  =\frac{4}{\sqrt{29}}\)

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,

\(cos\gamma  =\frac{\left<-3,4,2 \right>\cdot \left<0,0,1 \right>}{\sqrt{(-3)^{2}+4^{2}+2^{2}}\sqrt{0+0+1^{2}}}\)

\(cos\gamma   =\frac{-3(0)+4(0)+2(1)}{\sqrt{9+16+4}\sqrt{1}}\)

\(cos\gamma   =\frac{0+0+2}{\sqrt{29}\sqrt{1}}\)

\(cos\gamma  =\frac{2}{\sqrt{29}}\)

Step 2. Find the direction angles

Now, the direction angles are,

\(cos\alpha =\frac{-3}{\sqrt{29}}, \alpha =cos^{-1}\left ( \frac{-3}{\sqrt{29}} \right )\)

\(cos\beta =\frac{4}{\sqrt{29}}, \beta=cos^{-1}\left ( \frac{4}{\sqrt{29}} \right )\) 

\(cos\gamma =\frac{2}{\sqrt{29}}, \gamma=cos^{-1}\left ( \frac{2}{\sqrt{29}} \right )\)