### Video Transcript

Given that π¨ equals three π’ hat plus π£ hat plus ππ€ hat and that π© is a unit vector equal to one-fifth times π¨, determine the possible values of π.

To find what weβre looking for, weβll need to know what defines a unit vector and also how to perform the scalar multiplication one-fifth times π¨. The defining characteristic, the unit vector, as suggested by the word unit in its name, is that a unit vector has a magnitude of one. What this means for us is that the vector π© defined as one-fifth times π¨ has a magnitude of one. We represent magnitude using vertical bars and recall that the magnitude of a vector is the square root of the sum of the squares of its components. In this formula, π sub π is the πth component of the vector π, where we typically take the π’ hat component of a vector to be the first component, the π£ hat component to be the second component, and the π€ hat component to be the third component.

Now to multiply a vector by a scalar β in this case, π is the scalar and π is the vector β we simply multiply each component by that scalar. Using our previous equation for the magnitude of a vector, we can also come up with a very useful relationship for the magnitude of the scaled vector. If π is a scalar, then the magnitude of the vector π times π is equal to the absolute value of π times the magnitude of π. Algebraically, this makes sense because in our formula for the magnitude, ππ becomes π times ππ. So ππ squared becomes π squared times ππ squared. But when we factor out this common factor of π squared from each term in the sum, we get the square root of π squared, which is the absolute value of π, times our previous formula for the magnitude of π.

Geometrically, this makes sense because π scales the vector. So two times the vector should have twice the length, whereas one-half times the vector should have one-half times the length. We use the absolute value of π because, for negative numbers, the direction of the vector will also change in addition to its magnitude. But the magnitude of the vector is independent of its direction. Anyway, it is this relationship that is particularly useful to us. Instead of carrying out the scalar multiplication to find the vector one-fifth times π¨ and then applying the magnitude formula, we can instead find the magnitude of π¨ and then multiply by one-fifth.

So the magnitude of one-fifth times π¨ is one-fifth times the magnitude of π¨. And to calculate the magnitude of π¨ itself, we take the square root of the sum of the squares of its components. From our statement, we see that the components are three, positive one, and π. So the magnitude of π¨ is the square root of three squared plus one squared plus π squared. Three squared plus one squared is nine plus one, which is 10. We now have three results to combine that will give us a single equation for the unknown π. We know that one is equal to the magnitude of one-fifth times π¨ and that the magnitude of one-fifth times π¨ is one-fifth times the magnitude of π¨ and that the magnitude of π¨ is the square root of 10 plus π squared.

Putting this all together, we have one-fifth times the square root of 10 plus π squared equals one, where, again, the left-hand side is just an expansion of the magnitude of one-fifth times π¨. To solve this equation for π, weβll start by multiplying both sides by five. Now to eliminate the square root, weβll square both sides. And this gives us 10 plus π squared equals 25. Now, technically, we just took the square of a square root. So the left-hand side of this equation should really be the absolute value of 10 plus π squared.

Observe, though, that since π is real, π squared is always greater than or equal to zero. So 10 plus π squared is always positive, and specifying that this is an absolute value is redundant. To finish solving this equation, we subtract 10 from both sides. This gives us π squared equals 15. At this point, we would like to conclude that π is the square root of 15. But we have to be a little bit careful with the sign in this value.

Remember that π squared equals 15 actually has two solutions, positive square root 15 and negative square root 15. Thereβs nothing in the statement that restricts π to be positive or negative. Indeed, vector components can be positive, negative, or zero. In fact, one of the important consequences of the way we calculate the magnitude of a vector is that the magnitude of the vector turns out to be independent of the sign of its components.

All of this is to say that both positive square root 15 and negative square root 15 are possible values for π. And as our calculations have shown, they are the only values for π.