Vector 𝐁 has x, y, and z components of 4.00,6.00, and 3.00...

Vector $\overrightarrow{\mathbf{B}}$ has $x, y,$ and $z$ components of $4.00,6.00,$ and 3.00 units, respectively. Calculate (a) the magnitude of $\overrightarrow{\mathbf{B}}$ and (b) the angle that $\overrightarrow{\mathbf{B}}$ makes with each coordinate axis.

Key Concepts

Vector Components

This concept involves representing a vector by its projections along the coordinate axes. In three-dimensional space, a vector is expressed in terms of its x, y, and z components, which specify its magnitude in each direction and provide a complete description of its direction and length.

Magnitude of a Vector

The magnitude of a vector, often called its norm or length, is a measure of its overall size and is calculated using the Pythagorean theorem. In three dimensions, it is computed by taking the square root of the sum of the squares of its components. This measure is crucial for quantifying the extent or strength of the vector.

Direction Cosines

Direction cosines are the cosines of the angles between a vector and the coordinate axes. They provide a method to describe the orientation of the vector in space. By dividing each component by the magnitude, one obtains the cosine of the angle that the vector makes with each corresponding axis, thus connecting the vector's components with its geometric direction.

Transcript

00:01 Okay, in this problem, we're given a vector, vector b, and we're given its components.

00:07 The x component is 4 .00, we'll call that i -hat, plus the y component of 6 .00, we'll call that j -hat, and then plus 3 .00k hat.

00:21 And those are in units.

00:23 And first thing it wants this to do is find the magnitude of b, and the magnitude of b is just going to be each of the, e, three components squared.

00:34 So four squared plus six squared plus three squared.

00:40 And then you take the square root of it.

00:41 So it's kind of a super pythagoras thing.

00:44 And so that's going to be square root of 61.

00:49 And so the magnitude of b is equal to 7 .81 units.

00:57 And we'll be using that here a little bit in the next piece.

01:00 Because the next thing, and it's a really fascinating question, it's what angle does this vector make with each of the four axes? so what angle does that make with the x -axis, the y -axis, and the z -axis? and in order to solve that, we kind of need to remember the definition of what the dot product is.

01:23 And a dot b is equal to the magnitude of a times the magnitude of b times the cosine of theta.

01:32 And we can use this equation to find angle between any two vectors as long as we know what the dot product is.

01:42 And so if i solve for cosine theta, cosine theta is going to be a.

01:47 Dot b divided by the two magnitudes.

01:52 And so my theta is going to be equal to the inverse cosine of a dot b over magnitude of a and magnitude of b.

02:04 And we'll use that to solve for the three angles.

02:09 And so first, let's look at the x -axis.

02:12 Well, we have our two angles, and i'm going to write it in bracket form.

02:19 I think it's a little bit easier to process.

02:21 We've got b, which is 4, 6, and 3.

02:28 B is that one, x is that one.

02:30 And x is 1 -0...