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Question [2 points]Given the vectors u = <-14,-31 ,- 23> 0 = <-3,-2,-1> and w = <1,5,4> express as the sum of scalar multiples of v and w, e.g: U_...

Question

Question [2 points]Given the vectors u = <-14,-31 ,- 23> 0 = <-3,-2,-1> and w = <1,5,4> express as the sum of scalar multiples of v and w, e.g: U_Av+bw.WorksheetQuestion 2 [3 points]Given the vectors<-1,-7,2> ad 0 = <-3,-3,2> expresssum Of two vectors w1 and w2, where w1 is parallel to and w2 perpendicular to v.WorksheetW1WorksheetW2

Question [2 points] Given the vectors u = <-14,-31 ,- 23> 0 = <-3,-2,-1> and w = <1,5,4> express as the sum of scalar multiples of v and w, e.g: U_Av+bw. Worksheet Question 2 [3 points] Given the vectors <-1,-7,2> ad 0 = <-3,-3,2> express sum Of two vectors w1 and w2, where w1 is parallel to and w2 perpendicular to v. Worksheet W1 Worksheet W2



Answers

Let $\mathbf{u}, \mathbf{v},$ and $\mathbf{w}$ be the vectors in Exercise $11 .$ Find the components of the vector $x$ that satisfies the equation $3 \mathbf{u}+\mathbf{v}-2 \mathbf{w}=3 \mathbf{x}+2 \mathbf{w}$.

So if we want to show the vector negative to you plus three V where U is equal to zero negative 21 and V busy too. One negative one and zero. If we want to show that in component, form and unit vectors, we should put those into each other first to get negative to you. Plus three V is equal to negative two zero negative 21 plus three, one negative one and zero That would be equal to zero plus three, four minus three and to which is then equal to 31 and negative too. And this would be an answer. But then, if we want to put it into unit vectors, we can just kind of pull our normal unit vectors into here and call this three I plus J minus two K, and you can see how all of those are mapped directly onto there. You

In the problem, we have to find A across B. So this is written as i j k, it is minus two, five minus two, three minus two, and for this is equal to i m 2, 20 minus four minus jane to minus eight plus six plus, get into four minus 15. This is equal to 16, I plus two, j minus 11 K. Or this is recognized 16 two and minus 11, so this isn't the answer.

In the problem we have to find a cross and P. So this is written as I G. Okay, here is one, 23 minus two to minus four. This equals I into minus eight minus six minus J, into minus four plus six plus ken +22 plus two. Therefore this equals two minus 14, I minus two, J plus four K. Or this is written us. This is K. This is okay now this equals two minus 14 minus two and four. So overall this is the answer to the problem.

This question gives us two vectors and asks us to find a vector that's perpendicular to both of them. We have a special tool for this, and that's gonna be the cross product. We know that when we have two vectors a cross be that's always gonna be perpendicular to a and perpendicular to be. This was shown in your book by calculating a cross be dot product. A. The dot product, when two vectors are perpendicular, is always gonna be equal to zero. And we proved that no matter what A and B are a crust be dot a will equal zero. The same is true for be so a Crosby's perpendicular both two A and two B So let's jump into it the vectors that were given our A, which is equal to three j plus five k. Now, this is a linear combination of are unit vectors, which we can pretty easily turn into Just one factor. It's gonna be 035 The second vector that were given is B, which is equal to negative. I plus tu que again. That's gonna be negative. 10 and two. So let's start doing this. We know that to take the cross product. We're going to be, um, finding the determined of a three by three matrix whose first row, our unit vectors I j k are third and fourth or second and third row will be our vectors. So 035 and negative 102 So let's start. We know that taking the determined of a three by three matrix we're gonna do co factor decomposition. And I'm gonna use the first row because it's gonna be far easier than anything else. So what we'll have is the determinant of minor 11 which is just the vector or the Matrix without wrote one and column one 305 to multiple that might I that we're going to subtract, Remember, we're subtracting because of where J is in the Matrix because Element 12 and wanted to add up to an odd number. We're gonna have to subtract it. So we're subtracting minor 12 which is zero negative. One, 52 times j. And then we're adding again or adding minor 13 which is zero negative. 130 times k R unit Vector K. So now we can do this two by two Major sees taking the determiner of them is a breeze where six of minus zero, which is six. I we're gonna have zero minus negative five, which is positive. Five. Remember, we're subtracting it, so it's gonna be minus five J and then we're going to, uh, zero minus negative three, which is just gonna be positive. Three. So plus three. Okay. And this is our cross product. This is our resulting vector from his to, uh, original factors. So the vector six i minus five j plus three k, or in other words, six negative +53 is a vector that's gonna be perpendicular to both a and B. This isn't answered a part of the question. So if we have better a here and vector be here, our vector is something like this. Is it a cross be and we can see that it is at a right angle with a end with B. But hold on. We also want to find a unit vector that is perpendicular to both of these. Well, how can we do that? We're gonna have a special tool, which is not really a secret. Is just scaling this cross product no matter what the length of this cross product is, it's always going to be perpendicular. So we could double its length, give half its length. We could divide its length by pie, no matter what the vectors, and be perpendicular to A and B. So we want a unified A unit vector has a magnitude of one. What we want to do is to get a magnitude of one. We just want to divide or scale this vector down by its own magnitude. The magnitude right now is this sum of the square root of the some of each component squared. So six squared plus negative five squared plus three squared. When we square all of those add them up. What we get is the square root of 70. So right now this factor has elect of route 70 we wanted to have a length of one. So all we have to do is divide it by its length. So our unit vector is gonna be one over squared of 70 times six negative, five three. So this is a unit vector that is perpendicular to a and B and that's your final answer


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