Question

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

$⟨-1,1,-4⟩$

Short answer

The direction angles and direction cosines for the vectors are

$cos\alpha =\frac{-1}{\sqrt{18}},\alpha =co{s}^{-1}\left(\frac{-1}{\sqrt{18}}\right)$

$cos\beta =\frac{1}{\sqrt{18}},\beta =co{s}^{-1}\left(\frac{1}{\sqrt{18}}\right)$

$cos\gamma =\frac{-4}{\sqrt{18}},\gamma =co{s}^{-1}\left(\frac{-4}{\sqrt{18}}\right)$

Step-by-step solution

Step 1. Find the direction cosines

To find the direction cosines we will use the formula $cos\theta =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$.

Let the vector $u=⟨-1,1,-4⟩$.

So, the angle between the x-axis and the vector $u$ is $cos\alpha =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$ and $v=⟨1,0,0⟩$.

Therefore,

$cos\alpha =\frac{⟨-1,1,-4⟩\cdot ⟨1,0,0⟩}{\sqrt{(-1{)}^2+1^2+(-4{)}^2}\sqrt{1^2+0+0}}$

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

$cos\alpha =\frac{-1+0+0}{\sqrt{18}\sqrt{1}}$

$cos\alpha =\frac{-1}{\sqrt{18}}$

Now, the angle between the y-axis and the vector $u$ is $cos\beta =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$ and $v=⟨0,1,0⟩$.

Therefore,

$cos\beta =\frac{⟨-1,1,-4⟩\cdot ⟨0,1,0⟩}{\sqrt{(-1{)}^2+1^2+(-4{)}^2}\sqrt{0+1^2+0}}$

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

$cos\beta =\frac{0+1+0}{\sqrt{18}\sqrt{1}}$

$cos\beta =\frac{1}{\sqrt{18}}$

Now, the angle between the z-axis and the vector $u$ is $cos\gamma =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$ and $v=⟨0,0,1⟩$.

Therefore,

$cos\gamma =\frac{⟨-1,1,-4⟩\cdot ⟨0,0,1⟩}{\sqrt{(-1{)}^2+1^2+(-4{)}^2}\sqrt{0+0+1^2}}$

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

$cos\gamma =\frac{0+0-4}{\sqrt{18}\sqrt{1}}$

$cos\gamma =\frac{-4}{\sqrt{18}}$

Step 2. Find the direction angles

Now, the direction angles are,

$cos\alpha =\frac{-1}{\sqrt{18}},\alpha =co{s}^{-1}\left(\frac{-1}{\sqrt{18}}\right)$

$cos\beta =\frac{1}{\sqrt{18}},\beta =co{s}^{-1}\left(\frac{1}{\sqrt{18}}\right)$

$cos\gamma =\frac{-4}{\sqrt{18}},\gamma =co{s}^{-1}\left(\frac{-4}{\sqrt{18}}\right)$