Video: Using Operations on Vectors to Express a Vector in Terms of Two Other Vectors

Given that 𝐀 = βŒ©βˆ’4, βˆ’1βŒͺ and 𝐁 = βŒ©βˆ’2, βˆ’1βŒͺ, express 𝐂 = βŒ©βˆ’8, βˆ’1βŒͺ in terms of 𝐀 and 𝐁.

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Video Transcript

Given that the vector 𝐀 is equal to negative four, negative one and the vector 𝐁 is equal to negative two, negative one, express the vector 𝐂 which is equal to negative eight, negative one in terms of the vectors 𝐀 and 𝐁.

Here, we’ve been asked to express one vector, 𝐂, in terms of two others, vectors 𝐀 and 𝐁. Which means we’re looking to be able to write 𝐂 as π‘šπ€ plus π‘˜π for some scalar values π‘š and π‘˜. That is, some multiple of 𝐀 plus some multiple of 𝐁. We can replace 𝐀, 𝐁, and 𝐂 with their components, and we have a vector equation. The vector negative eight, negative one is equal to π‘š multiplied by the vector negative four, negative one plus π‘˜ multiplied by the vector negative two, negative one. We then recall that in order to multiply a vector by a scalar, we just multiply each of its components by that scalar.

So, to multiply the vector negative four, negative one by π‘š, we first multiply π‘š by negative four, giving a first component of negative four π‘š. And then, we multiply π‘š by negative one, giving a second component of negative π‘š. In the same way, to multiply the vector negative two, negative one by π‘˜, we first multiply π‘˜ by negative two, giving a first component of negative two π‘˜. And then, we multiply π‘˜ by negative one, giving a second component of negative π‘˜. Next, we need to add the two vectors in the right-hand side together. And to do this, we recall that we can simply add the component parts.

So, the first component of the vector on the right-hand side will be the sum of the first components. That’s negative four π‘š plus negative two π‘˜, which we can write as negative four π‘š minus two π‘˜. And then, the second component will be the sum of the second component in each vector. That’s negative π‘š plus negative π‘˜, which we can write as negative π‘š minus π‘˜. So, we now have that the vector negative eight, negative one is equal to the vector negative four π‘š minus two π‘˜, negative π‘š minus π‘˜. At this point, we recall that if two vectors are equal, then it must mean that their individual components are equal.

So, the first component of the vector on the left-hand side must be equal to the first component on the right-hand side. This gives us an equation without any vectors. Negative eight is equal to negative four π‘š minus two π‘˜. In the same way, equating the second components in our vectors gives the equation negative one equals negative π‘š minus π‘˜. We now have a pair of linear simultaneous equations in the scalars π‘š and π‘˜. So, we can solve these simultaneous equations to find their values.

First, though, we can simplify each equation slightly. The first equation can be divided by negative two to give the simplified equation two π‘š plus π‘˜ equals four. And the second equation can be divided by negative one to give the simplified equation π‘š plus π‘˜ equals one. We then see that we have the same coefficient of π‘˜ in each of our equations. So, if we subtract the second equation from the first, this will eliminate the π‘˜ terms. On the left-hand side, two π‘š plus π‘˜ minus π‘š minus π‘˜ will leave us with π‘š. And on the right-hand side, four minus one is equal to three. So, in fact, we found the value of π‘š. π‘š is equal to three.

We can then substitute this value of π‘š into our simplest equation, which is the equation π‘š plus π‘˜ is equal to one. So, we have that three plus π‘˜ is equal to one. And this equation can be solved by subtracting three from each side to give π‘˜ is equal to negative two. We’ve therefore found the values of our two scalars π‘š and π‘˜. Finally, we can return to the statement we began with, which was that the vector 𝐂 is equal to some scalar multiplied by 𝐀 plus some scalar multiplied by 𝐁 and substitute the values of π‘š and π‘˜.

We found then that the vector 𝐂 can be expressed in terms of the vectors 𝐀 and 𝐁 as 𝐂 equals three 𝐀 minus two 𝐁. We can, of course, check this manually by recalculating the vector three 𝐀 minus two 𝐁. And if we do so, we find that we get the vector negative eight, negative one, which is indeed equal to the vector 𝐂. So, our answer is correct.

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