In Exercises 53–56, let
u = -2i + 3j, v = 6i - j, w = -3i.

Find ...

Question

In Exercises 53–56, let
u = -2i + 3j, v = 6i - j, w = -3i.

Find each specified vector or scalar.
4u - (2v - w)

Accepted Answer

Hey, everyone in this problem, we're asked to find seven A minus the binomial three B minus two C using the following vectors. So we have A, is equal to negative five, I plus six JB is equal to two, I plus seven J and C is equal to 10, I plus nine J. We have four answer choices and all of them have 21 I and 39 J. And they just differ in whether they have positive or negative for both of those I or J components. So let's start by writing out what we were given. We have seven A minus this factor with three B minus two C. So let's start by just filling in and substituting in the vectors we were given. OK. So this expression is gonna be seven multiplied by negative five I plus six J minus. We're gonna do big square bracket here because we have three multiplied by two, I plus seven J minus two, multiplied by 10 I plus nine J. So what we have here is scalar multiples OK of our vectors. And we're also adding and subtracting our vectors and we want to recall and we have a scalar multiple of a vector. What we're gonna do is just multiply that entire vector by that scale. OK. So each component of our vector is going to be multiplied. And when we're adding and subtracting vectors, we just do that like we regularly would component wise. OK? So we're gonna add up all the I components, we're gonna add up all the J components. So what that means is as we have it written here, what we need to do is expand and collect like terms like we would in any other case. OK. So let's go ahead and do that. In this first term. We have seven multiplied by negative five I plus six J. We're gonna apply that seven to both terms. Expand our bracket. We're gonna end up with negative 35 I plus 42 J and then we're gonna subtract in this big bracket and we're gonna expand all of this as well. Starting with our three. That's gonna apply to both two I and seven J. We get six I plus 21 J and now we're gonna subtract, yeah, we're gonna leave this in brackets. So we don't mix up our negative sign and we're gonna apply this two to both of the terms inside the bracket. 10 I and nine J, this gives us 20 I 18. All right. Now we have this negative in front of this brack here. So we have negative and then we have 20 I plus 18 J. And we know that, that negative needs to apply to everything inside of the bracket. So let's simplify within the bracket. First, negative 35 I plus 42 J minus six, I plus 21 J minus I minus 18 J. All right. So it's simplified within that bracket. Now, we can do the same again. Can we have negative this entire second half of our equation that negative has to apply to all those terms. So let's apply that same principle. We get negative 35 I J. My mistakes, I minus 21 J 4 20 I plus 18 J. OK. So those positive terms became negative, the negative terms became positive. And now we have this big long expression. We've gotten rid of all of the brackets. Everything's expanded as much as possible and we're left with just collecting like terms and simplifying. So we're gonna look for all of our I terms. We have negative 35 I negative six I 20 I adding these all up gives us negative 21 I and we're gonna do the same for our J terms. So we have 42 J negative 21 J J adding all those up give us positive 39 J. And so what we get in the end is the vector negative 21 I plus 39 J. And that is the result of the expression we were looking for if we compare this to our answer choices. We can see that this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 53–56, let u = -2i + 3j, v = 6i - j, w = -3i. Find each specified vector or scalar. 4u - (2v - w)

Key Concepts

Vector Operations

Scalar Multiplication

Vector Subtraction

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