Chapter 9, Problem 24CT

To determine

To calculate: The direction angle for $u$, where $u=2i-3j+k$ .

Answer to Problem 24CT

Solution:

The direction angles for $u=2i-3j+k$ are approximately $\alpha ={57.7}^0,\beta ={143.3}^0,\gamma ={74.5}^0$

Explanation of Solution

Given information:

The vector, $u=2i-3j+k$

Formula used:

1) The magnitude of vector $u=a_{1}i+a_{2}j+a_{3}k$ , $\left|\right|u\left|\right|=\sqrt{a_1{}^2+b_1{}^2+c_1{}^2}$ .

2) The direction angles of the vector $u=a_{1}i+a_{2}j+a_{3}k$ , are $\alpha ,\beta ,\gamma$ where $\cos \alpha =\frac{a_1}{\left|\right|u\left|\right|}$ ,

  $\cos \beta =\frac{a_2}{\left|\right|u\left|\right|}$ , $\cos \gamma =\frac{a_3}{\left|\right|u\left|\right|}$ .

Calculation:

Let $u=2i-3j+k$ .

By using $\left|\right|u\left|\right|=\sqrt{a_1{}^2+b_1{}^2+c_1{}^2}$ ,

  $\left|\right|u\left|\right|=\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(1\right)}^2}=\sqrt{4+9+1}=\sqrt{14}$.

Now consider, $\cos \alpha =\frac{a_1}{\left|\right|u\left|\right|}$ .

  $\Rightarrow \cos \alpha =\frac{2}{\sqrt{14}}$

  $\Rightarrow \cos \alpha =\frac{2\sqrt{14}}{14}$

Applying cos inverse on both sides,

  $\alpha ={\cos }^{-1}\left(\frac{2\sqrt{14}}{14}\right)$

  $\alpha ={57.7}^0$

Now consider $\cos \beta =\frac{a_2}{\left|\right|u\left|\right|}$ .

  $\Rightarrow \cos \beta =\frac{-3}{\sqrt{14}}$

  $\Rightarrow \cos \beta =\frac{-3\sqrt{14}}{14}$

Applying cos inverse on both sides,

  $\beta ={\cos }^{-1}\left(\frac{-3\sqrt{14}}{14}\right)$

  $\beta ={143.3}^0$

Consider $\cos \gamma =\frac{a_3}{\left|\right|u\left|\right|}$

  $\Rightarrow \cos \gamma =\frac{1}{\sqrt{14}}$

  $\Rightarrow \cos \gamma =\frac{\sqrt{14}}{14}$

Applying cos inverse on both sides,

  $\gamma ={\cos }^{-1}\left(\frac{\sqrt{14}}{14}\right)$

  $\gamma ={74.5}^0$

Therefore, the direction angles of the vector $u=2i-3j+k$ are $\alpha ={57.7}^0,\beta ={143.3}^0,\gamma ={74.5}^0$ .