Answer:
Step-by-step explanation:
given is a vector as (5,9,3)
a = (5,9,3)
To find out direction cosines
First let us calculate modulus of vector a
[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]
Direction ratios are (5,9,3)
Magnitude of vector a = [tex]\sqrt{115}[/tex]
So direction cosines would be
[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]
Angles would be
[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]
=cos inverse (0.4662, 0.8393, 0.2798)
= (62.21, 32.93,32,94)
