Video: Finding the Dot Product between Vectors

The angle between 𝐀 and 𝐁 is 22Β°. If |𝐀| = 3, |𝐁| = 25.2, find 𝐀 β‹… 𝐁 to the nearest hundredth.

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Video Transcript

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 πœƒ 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 on.

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.

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