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.
