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Solve the following problems:Find the angle between & = [1 - 2 1JT and & = [1 2 _ 7T4 Using the same vectors 6 and & as above, compute proja(v)_ Show us...

Question

Solve the following problems:Find the angle between & = [1 - 2 1JT and & = [1 2 _ 7T4 Using the same vectors 6 and & as above, compute proja(v)_ Show using the definition of projection and the algebraic properties of dot product, that for any two non-Zero vectors & and 6 you have that proja(6)) 6 = 0 This shows that any vector 6 in Rn can be written as the addition of a vector perpen- dicular to a given vector & plus a vector parallel to & (d) Write & = [1 _ 2 4T as t

Solve the following problems: Find the angle between & = [1 - 2 1JT and & = [1 2 _ 7T4 Using the same vectors 6 and & as above, compute proja(v)_ Show using the definition of projection and the algebraic properties of dot product, that for any two non-Zero vectors & and 6 you have that proja(6)) 6 = 0 This shows that any vector 6 in Rn can be written as the addition of a vector perpen- dicular to a given vector & plus a vector parallel to & (d) Write & = [1 _ 2 4T as the addition of two vectors. One of them should be perpendicular to & = [1 2 and the other should be parallel t0 to & = [1 2



Answers

For the vectors $\mathbf{a}, \mathbf{b},$ and $\mathbf{c}$ from Problem $6,$ find the angle between each pair of vectors.

So for this problem, we're given the vector You equal the ABC and were asked to show that thesis Kwehr Sze of all of the direction co signs equals one. So to do that, we're going to take this turn my term and right out and we'll show that this simplifies down. So first we're going to have the co sign of Hey, which is going to be a over square root of a squared plus b squared plus c squared. And then all of this will be squared and we're going to add that to be over once again. Hey, squared B squared c squared all that squared and finally see over a squared B squared, C squared, all squared And so this is going to be equal to ah, a squared plus B squared plus c squared on the top And the bottom of these will all be a squared plus b squared plus c squared, which means that this is indeed one. So we have showed this

Okay we're giving you is equal to one common Sue comma two. And we have to find the whole time angles. So it's do court saying of Alfa, this is with you and I And so yet one times one plus two times l plus 2 10 shops all over the magnitude of you, which is nine. And the magnitude of eye, which is one that's is equal to one over three. An Alfa is equal to actually no, you can keep your foot back cause we're trying to justify our I didn't have to take now coastline data that's going to be with you and Jay. So that's one time cell plus one times or two times one plus to tell yourself all over Scripture tonight. Excuse me to over three now, co sign of mule, which is with U N. Okay, Well, Jay had to and Kay has to too. So we're gonna get this evening. We're going good to over. Great. Now we're going to show that co sign squared out for a plus co sign Squared Veda plus co sign Squared Mute. Is he good to one? Well, let's see what's co sign squared Alfa. Well, That's one over three squares that one overnight plus coastline beta. It's two squared over three square. Instead, that's four overnight plus for over nine. So that's +456789 Do we get nine overnight and that is equal to one?

Let's find the dot product of two vectors if their length are six and one third, and the angle between them is Viagra. For so we know the dot product. It's gonna be the magnitude of one which is six times the magnitude of other, which is 1/3. I'm the co sign of the angle between them. So in this case were however, for which is in radiance, not degrees. So you end up getting this right here, which we know is the same thing as The Square Root of two. So route to, we'll end up being um our final answer for this problem for this whole section. We're going to be looking at the dot product and the ways in which we calculate the dot product because we know it's going to equal the magnitudes times the cosine of the angle between them. Um And in these following problems, we're going to actually calculate the dot product just given the factors.

Hello there. So this exercise we need to find a vector in R. two or a vector of two components with length equals to two. And that form an angle with the X axis equals two by fours. So well, the fact that these vector forms an angle of my first means that the component on the X. On the the X. Component and the white component will be the same. In other words we're going to form a triangle of this components. A. And If both sides are the same then these will form a pie 4th. And now we need that the length of this vector here uh Should be equal to two. So that means that the square root of these components, a square was a square should be equal institute and from this week we obtain this is equivalent To have that a. Is equal to the square root of two. Okay, so this gives us the components. So the vector itself will be the vector squared of two. I Plus the Square Root of two. Yeah. Right now we need to find another vector. Now this factory will be in our three director of three components. It will be a unitary vector. That means that the length of the vector is a close to one and it's located in the uh wise in the XZ plane. So here we have the Y axis, the X axis and easy access. And this victory will be in this part in the exit plane and will form an angle with respect to the to the X axis. It calls to ah Theatre equals two pi over six. Okay, so that gives us enough information to construct these vectors. So you need to re vector even an angle will be equal to just the co sign here of pie Over six because we are Mhm. We want this the length of this side of the component of this side of the factory. So that means taking just a co sign of design. So the go sign of 56 in the I direction plus the sign of pie sixth indicate direction because this this victory is located in the XZ plane and you can observe the design will give us the component in dizzy access. So by resolving this we obtain the vector is equal to the square root of three over to I plus one house. Okay?


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