Given that the magnitude of 𝐀 is
equal to the magnitude of 𝐁, 𝐀 dot 𝐁 is equal to 69, and the measure of the angle
between 𝐀 and 𝐁 is 60 degrees, determine the magnitude of 𝐀 to the nearest
We’re dealing with two vectors in
this question, vector 𝐀 and vector 𝐁. We’re told the angle between these
vectors is equal to 60 degrees. When we’re thinking about angles
between two vectors, we should be thinking about a scalar product. This says that if 𝐀 and 𝐁 are two
vectors, cos of 𝜃 is equal to 𝑎 dot 𝑏 over the magnitude of 𝑎 times the
magnitude of 𝑏.
Now in this case, 𝜃 is equal to
60. So we have cos of 60 degrees. We’re told that the dot product of
𝐀 and 𝐁 is equal to 69. Then this is all over the magnitude
of 𝐀 times the magnitude of 𝐁. Now of course, we know that cos of
60 degrees is equal to one-half. But we’re also told that the
magnitude of 𝐀 is equal to the magnitude of 𝐁. So we can rewrite the magnitude of
𝐀 times the magnitude of 𝐁 as the magnitude of 𝐀 squared.
To solve for this equation for the
magnitude of 𝐀, we’ll begin by multiplying both sides by the magnitude of 𝐀
squared. And we have one-half times the
magnitude of 𝐀 squared equals 69. Next, we’ll multiply through by
two. And we find the magnitude of 𝐀
squared to be equal to 138.
Our final step is going to be to
find the square root of both sides. Now normally, we would find both
the positive and negative square root of 138. But the magnitude must be a
positive value. So we’re only gonna find the
positive square root of 138. Well, the positive square root of
138 is 11.7473 and so on. Rounding this to the nearest one
hundredth, we find the magnitude of 𝐀 to be equal to 11.75.