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

If 𝐀 = ⟨1, βˆ’2, 2⟩, 𝐀 βˆ₯ 𝐁, |𝐁| = 6, and the direction of 𝐁 is opposite to that of 𝐀, find 𝐁.

02:39

Video Transcript

If 𝐀 is the vector one, negative two, two and the vector 𝐀 is parallel to the vector 𝐁 and the modulus of vector 𝐁 is equal to six, and the direction of 𝐁 is opposite to that of vector 𝐀, find vector 𝐁.

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

And it’s worth reiterating here our value of π‘˜ must be positive. So the direction of 𝐁 is opposite to that of vector 𝐀. 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 𝐁 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 𝐁 is equal to six.

And we want to use this to find an expression for the magnitude of 𝐁. And to do this, we recall the magnitude of a vector π‘Ž, 𝑏, 𝑐 is the square root of the sum of the squares of its components. That’s the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared. So if we apply this, we get six is equal to the magnitude of vector 𝐁, which is equal to the square root of the sum of the squares of its components. That’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 we’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 𝐁. This then gives us that 𝐁 is equal to the vector negative two, two times two, negative two multiplied by two. And we can evaluate each component together. 𝐁 is the vector negative two, four, negative four, which is our final answer. Therefore, we were able to show if 𝐀 is the vector one, negative two, two, the magnitude of vector 𝐁 is six, and the direction of vector 𝐁 is opposite to that of vector 𝐀, then 𝐁 must be the vector negative two, four, negative four.

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