Question

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1, 4

Answer 1

Answer:

The direction cosines are:

$\frac{5}{\sqrt{42} }$, $\frac{1}{\sqrt{42} }$  and  $\frac{4}{\sqrt{42} }$  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = $\frac{a . i}{|a| . |i|}$               ---------------------(i)

cos β = $\frac{a.j}{|a||j|}$               ---------------------(ii)

cos γ = $\frac{a.k}{|a|.|k|}$             ----------------------(iii)

And from these we can get the direction angles as follows;

α =  cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i is just the x component of the vector]

a . j = 1            [the y component of the vector]

a . k = 4          [the z component of the vector]

Also

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = $\sqrt{5^2 + 1^2 + 4^2}$

|a| = $\sqrt{25 + 1 + 16}$

|a| = $\sqrt{42}$

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = $\frac{5}{\sqrt{42} }$

cos β =  $\frac{1}{\sqrt{42} }$              

cos γ =  $\frac{4}{\sqrt{42} }$

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

α =  cos⁻¹ ( $\frac{5}{\sqrt{42} }$ )

α =  cos⁻¹ ($\frac{5}{6.481}$ )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

β = cos⁻¹ ( $\frac{1}{\sqrt{42} }$ )

β = cos⁻¹ ( $\frac{1}{6.481 }$ )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

γ = cos⁻¹ ($\frac{4}{\sqrt{42} }$)

γ = cos⁻¹ ($\frac{4}{6.481}$)

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

Conclusion:

The direction cosines are:

$\frac{5}{\sqrt{42} }$, $\frac{1}{\sqrt{42} }$  and  $\frac{4}{\sqrt{42} }$  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.