(a) Find the scalar product of the two vectors A⃗ and B⃗ given in Exercise 1.42 . (b) Find the a

(a) Find the scalar product of the two vectors A⃗ and B⃗...

Key Concepts

Scalar (Dot) Product

The scalar, or dot, product is an operation in vector algebra that takes two vectors and returns a scalar by multiplying corresponding components and summing the results. Alternatively, it can be computed using the magnitudes of the vectors and the cosine of the angle between them. This operation is fundamental in determining projections, checking for orthogonality, and computing work done by a force in physics.

Angle Between Vectors

The angle between two vectors is found by using the relationship between the dot product and the magnitudes of the vectors. By dividing the dot product by the product of the magnitudes, one obtains the cosine of the angle between the vectors. The inverse cosine operation can then be used to determine the actual angle. This concept is key in understanding the orientation and directional relationships between vectors in space.

Transcript

00:01 For this question, we're asked to use vector a and vector b given an exercise 1 .42, and i wrote those out of the top here.

00:07 I wrote the i -hat and j -hack components of both vector a and vector b.

00:10 For part a, it says find the scalar product of the two vectors, and part b wants us to find the magnitude between the two vectors.

00:17 So for part a, the scalar product, or also known as the dot product, is going to be a .b.

00:25 So this is equal to the i -hack components of a multiplied by the i -hack.

00:32 Components of b, so this is going to be 4 .0 multiplied by 5 .0, plus the j -hack components of a times the j -hack components of b.

00:44 So this is going to be 7 .0 times, and the negative numbers matter.

00:49 So this is going to be times negative 2 .0.

00:52 And it gives us a scalar.

00:54 That's why they call it a scalar product, and a scalar is just a number, meaning it doesn't have any i -had or j -hat values associated with it.

00:59 So we have 4 .0 times 5 .0 plus 7 .0 times negative of 2 .0.

01:05 Or in other words, 20 minus 14, which comes out to be 6, or 6 .0.

01:12 And we can box that in as a solution for a.

01:14 Part b wants us to find the angle between the two.

01:17 Well, another way of calculating the scalar product, a .b, is also by taking the magnitude of a, multiplying it by the magnitude of b, and then multiplying that by the cosine of the angle in between them.

01:36 We're going to call that angle theta a -b.

01:37 And that's what we're trying to find is theta ab...

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