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

Question

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

Accepted Answer

Hey, everyone. Welcome back in this problem. We're asked to find the magnitude of A plus B squared minus the magnitude of A minus B squared using the vectors A is equal to negative five I plus the J and B is equal to two I plus seven J. OK? And just a reminder that those double lines there around our A plus B and our A minus B, those indicate the magnitude. Now we have four answer choices for this problem. Option A 228 option B negative 10 I plus 12 J. Option C negative 128 and option D positive 128 right off the bat. We can already go ahead and eliminate option B because option B is a vector. But our expression is the subtraction of two magnitudes. OK? We know the magnitude is gonna be just a number. When we subtract two numbers, we're gonna get just a number. OK? So we are not gonna get a vector out of this expression. So we can already eliminate answer choice B. Now let's write out what we're looking for. So if the magnitude of A plus B squared minus the magnitude of A minus B squared. Then let's start just by substituting in our vectors A and B and, and this is gonna be equal to the magnitude of the vector A negative five I plus six J plus plus B which is two I plus seven J. And that's all squared. The numbers are tracking the magnitude of a negative five I left six J minus B two I 47 J. Then again, all squared. All right. So this looks like a big messy expression. Let's go step by step and simplify inside of our magnitudes. May we have either the addition or the subtraction of two vectors? You recall that when we add or subtract vectors, we're just gonna add the corresponding components. OK? So we're gonna kind of group and combine like terms here. So inside this first magnitude, OK, we're adding these two vectors. We're gonna start with the I components. So we have negative five I and we have a positive two I. Well, that's gonna give us negative three. I, we're gonna do the same for the J component. We have plus six J um plus seven J. So it's gonna give us plus 13 J and this entire thing is squared. So now you just have this single vector inside of our magnitude off. Now, for the second one, OK, we're subtracting this one, we have the magnitude and now we have a subtraction of vector. And so that negative sign we have has to apply to both terms in our vector. So what we can do is to expand these brackets first just to make it a little bit more straightforward. And then we can simplify it. So we're gonna have negative five I plus six J minus two I minus seven J again, expanding that negative into the brackets, all of this is squared. And now we can collect like terms and simplify in that second magnitude expression. So we get the magnitude of negative three I plus J, all squared minus the magnitude of negative five I minus two. I gives us negative seven I six J minus seven J gives us minus J. And again, all squared. Now we have these magnitude square. OK. So let's recall that the magnitude of a vector, if we write our vector, let me write this in a different color just so it stands out. So we write our vector as I that's YJ. The magnitude of this vector is gonna be the square root of X squared plus Y squared. So it's the square root of the sum of the squares of each component. Now we want the magnitude squared, so we have the square root and then we're gonna swear it. And so this is just gonna give us X squared plus Y squared. All right. So we can simplify our expression. First, we have the magnitude squared of negative three I plus 13 J. Now we're gonna take that first component negative three and square it. And then we're gonna add the second component 13 and square it and we're gonna subtract and we're gonna subtract the same thing from the second expression. We have the first component negative seven squared plus the second component negative one squared. And be careful with your brackets and your signs here because we're subtracting this entire second magnitude squared. Both our negative seven squared and our negative one squared need to be inside of this bracket that minus is gonna apply to both of those. All right. And now we're done with the magnitudes. We're done with the vectors. We just have this expression of numbers that we need to simplify. So we get nine 4 to minus 49 4th one. If we work this all out, we get that, the expression we were given is gonna be equal to 128. And this corresponds with answer choice. And D 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. ||u + v||² - ||u - v||²

Key Concepts

Vector Addition and Subtraction

Vector Norm (Magnitude)

Properties of Dot Product and Norms

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