Question Video: Finding a Vector given Its Norm and Direction Angles Mathematics

Video Transcript

Find vector 𝐀 whose norm is 41 and whose direction angles are 135 degrees, 120 degrees, and 60 degrees.

We know that the norm of a vector is its magnitude and that the direction vectors 𝛼, 𝛽, and 𝛾 are the angles between the unit vectors 𝐢, 𝐣, and 𝐤 and the vector 𝐯. We also know that the direction cosines are such that cos 𝛼 is equal to 𝐯 sub 𝑥 over the magnitude of 𝐯, cos 𝛽 is equal to 𝐯 sub 𝑦 over the magnitude of 𝐯, and cos 𝛾 is equal to 𝐯 sub 𝑧 over the magnitude of 𝐯, where the vector 𝐯 has components 𝐯 sub 𝑥, 𝐯 sub 𝑦, and 𝐯 sub 𝑧. Substituting in our values for 𝛼 and the magnitude, we see that the cos of 135 degrees is equal to 𝐀 sub 𝑥 over 41. We can then multiply both sides of this equation by 41. 𝐀 sub 𝑥 is equal to negative 41 root two over two. This is the 𝑥-component of our vector 𝐀.

In the same way, considering angle 𝛽, we have the cos of 120 degrees is equal to 𝐀 sub 𝑦 over 41. Once again, we multiply both sides of the equation by 41 such that 𝐀 sub 𝑦 is equal to negative 41 over two. Finally, we have the cos of 60 degrees, angle 𝛾, is equal to 𝐀 sub 𝑧 divided by 41. 𝐀 sub 𝑧 is therefore equal to 41 multiplied by the cos of 60 degrees, which in turn is equal to 41 over two. This means that vector 𝐀 has components negative 41 root two over two, negative 41 over two, and 41 over two. This is the vector whose magnitude or norm is 41 and direction angles are 135, 120, 60 degrees.