The angle between vector 𝐀 and
vector 𝐁 is equal to 22 degrees. If the magnitude of vector 𝐀
equals three and the magnitude of vector 𝐁 equals 25.2, find the dot product of
vector 𝐀 and 𝐁 to the nearest hundredth.
When dealing with the angle between
two vectors, we recall that the cos of this angle 𝜃 is equal to the dot product of
the two vectors divided by the product of the magnitude of the two vectors. In this question, we’re told that
the angle is 22 degrees. The magnitude of vector 𝐀 is equal
to three. The magnitude of vector 𝐁 is equal
to 25.2. We need to work out the value of
the dot product of vector 𝐀 and vector 𝐁.
Substituting in our values gives us
cos of 22 degrees is equal to the dot product of vectors 𝐀 and 𝐁 divided by three
multiplied by 25.2. Three multiplied by 25.2 is equal
to 75.6. We can then multiply both sides of
the equation by 75.6. The dot product of vectors 𝐀 and
𝐁 is equal to 75.6 multiplied by the cos of 22 degrees. This is equal to 70.0950 and so
We have been asked to round our
answer to the nearest hundredth. This is the same as rounding to two
decimal places. As our deciding number is a five,
we round up. Therefore, the dot product of
vectors 𝐀 and 𝐁 is 70.10.