If the magnitude of 𝐀 equals five,
the magnitude of 𝐁 equals 15, and the measure of the angle between them is 30
degrees, find the algebraic projection of 𝐁 in the direction of 𝐀.
Alright, so in this exercise, we
have these two vectors 𝐀 and 𝐁. And we know that the magnitude of
𝐀 is one-third the magnitude of 𝐁 and also that they’re separated by an angle of
30 degrees. Knowing this, we want to solve for
the algebraic projection, also known as the scalar projection, of vector 𝐁 in the
direction of vector 𝐀. What we’re proposing then is
effectively to lay vector 𝐁 down along this line we’ve drawn through vector 𝐀. Essentially, we’re solving for how
much of vector 𝐁 is parallel to vector 𝐀.
We can figure out how to compute
this scalar projection in one of two ways. The first way is to recall from
memory that the scalar projection of one vector, we’ve called it 𝐕 one, onto
another, 𝐕 two, is given by these two expressions here. So, for example, we could use the
fact that the magnitude of the first vector multiplied by the cosine of the angle
between the two vectors gives us this scalar projection to solve for the algebraic
projection of 𝐁 onto 𝐀.
Notice though that we could also
reach this conclusion based on our sketch of vectors 𝐀 and 𝐁. The vector 𝐁 essentially forms the
hypotenuse of a right triangle. And so we see that it’s the
magnitude of that hypotenuse, the magnitude of 𝐁, multiplied by the cosine of the
angle between 𝐁 and the direction of 𝐀, which is 30 degrees, that gives us this
algebraic projection. Either method we choose leads us to
the same result.
We substitute in the magnitude of
𝐁, 15. And knowing that the cos of 30
degrees equals exactly the square root of three over two, we find our answer to be
15 times the square root of three over two. This is the algebraic or scalar
projection of 𝐁 in the direction of 𝐀.