# Question Video: Finding a Vector from Its Magnitude and Relationship with a Given Vector Mathematics

If π = β¨1, β2, 2β©, π β₯ π, |π| = 6, and the direction of π is opposite to that of π, find π.

02:39

### Video Transcript

If π is the vector one, negative two, two and the vector π is parallel to the vector π and the modulus of vector π is equal to six, and the direction of π is opposite to that of vector π, find vector π.

In this question, weβre given some information about two vectors, vectors π and π. We need to use this information to determine vector π. To do this, we start by noting that vectors π and π in parallel and in particular the direction of π is opposite to that of vector π. So this means that vector π must be some negative scalar multiple of π. In other words, π is equal to negative ππ. And weβre given vector π. π is the vector one, negative two, two. So we can substitute this into our equation. So π is equal to negative π multiplied by the vector one, negative two, two.

And itβs worth reiterating here our value of π must be positive. So the direction of π is opposite to that of vector π. We can now evaluate the scalar multiplication by multiplying all of the components of the vector by the scalar. This then gives us the vector negative π, two π, negative two π. So we found an expression for vector π in terms of π. However, we need to find the value of π. And to do this, we need to use the fact that the magnitude of vector π is equal to six.

And we want to use this to find an expression for the magnitude of π. And to do this, we recall the magnitude of a vector π, π, π is the square root of the sum of the squares of its components. Thatβs the square root of π squared plus π squared plus π squared. So if we apply this, we get six is equal to the magnitude of vector π, which is equal to the square root of the sum of the squares of its components. Thatβs root negative π squared plus two π all squared plus negative two π all squared.

And now, we can evaluate this. We get the square root of π squared plus four π squared plus four π squared, which simplifies to give us root nine π squared. And remember, we want the positive square root, and π is positive. So this simplifies to give us three π. So weβve shown that six is equal to three π. We can divide both sides of the equation through by three to see that π is equal to two.

Now, we just need to substitute π is equal to two into our expression for the vector π. This then gives us that π is equal to the vector negative two, two times two, negative two multiplied by two. And we can evaluate each component together. π is the vector negative two, four, negative four, which is our final answer. Therefore, we were able to show if π is the vector one, negative two, two, the magnitude of vector π is six, and the direction of vector π is opposite to that of vector π, then π must be the vector negative two, four, negative four.