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13.points) Find three vectors orthogonal to~214,points) Find four vectors orthogonal to15. (5 points) Let 7 =() eR?. Find a vector e R3 for which 7 +3 =...

Question

13.points) Find three vectors orthogonal to~214,points) Find four vectors orthogonal to15. (5 points) Let 7 =() eR?. Find a vector e R3 for which 7 +3 =

13. points) Find three vectors orthogonal to ~2 14, points) Find four vectors orthogonal to 15. (5 points) Let 7 = () eR?. Find a vector e R3 for which 7 +3 =



Answers

Find a vector that is perpendicular to the plane passing through the three given points. $$ P(1,1,-5), Q(2,2,0), R(0,0,0) $$

In this problem, we are given three points and asked to find or to produce a vector that is perpendicular to the plane that contains those three points. And to do this, we're going to take the vector p Q. And take the cross product of that vector with the vector p R. So how do we get P Q and P. R. Well, to get a peek, you were going to take the point Q. And subtract the point p from it. That's because Q is the end point of the vector and P is the starting point. And so we're going to get one 23 minus three 45 And our factor is going to be the foaming we get negative, too negative, too negative, too. Okay. Now, to get the vector PR, we're going to take the point are and subtract the point p from it. And so we're going to get four seven six, minus three for five. And when we do that, we get the following factor. One, three, one And again, that is equal to the vector. PR for the one of here is equal to the vector p Q. And now we're going to take the cross product of these two doctors by running them in a matrix that has the first row of I, J and K a second row with the components of P Q, which are native to negative to negative two and 1/3 row with the components of PR, which are 13 in one. Okay, and now to take the cross product and get the component for I, we're going to cross out the first column and take the determinant of this two by two matrix when we get negative. Two times one minus native, two times three as native to minus native six United to plus six, which is for then, we always attracts the J component to get back up or not, we're going to cross at the middle column and take the determinant of this to go to Matrix. So we get negative two times one minus negative two times one her native to plus two zero and then to get the K component, we're going to cross out the third calm. You take the determinant of this two by two matrix, which is going to be negative two times three minus data two times one which is negative. Six minus negative to Grenada, six plus two or negative for and we have found our vector. It's going to be four zero negative for.

So we're given three points p. Q and R. And our task is to find the vector perpendicular to the plane that contains thes three points. The way we're going to do this is first by developing our vectors P Q and P. R. So are two vectors are p Q and P. R. So peek you is going to be equal to Well, we know P is our initial point. Q. Is Air terminal point, so we have to do each of the elements of Q minus each of the elements of Pete. So that's gonna be one minus three comma to minus four, comma three minus five. And just a reminder we did it in this order because 12312123123 are all the terminal elements in this peak. You point vector calculation because Q is the terminal point in the vector. P is the initial point, which means they're the second terms that come in these subtraction components. In PR, we repeat the same process, using our instead of cute as our terminal points. So cue so are the way we're gonna use it is in the same way. Four minus three seven minus four and six minus five. Now we can solve both of these and we get our Spectres as negative two. Negative, too. I think it's two and the negative, too. And for a 2nd 1 we get 131 No. All we have to do to find the vector perpendicular to the plan containing these three points is we have to find the cross product peak. You cross PR p Q Cross P. R. So let's jog her memory on the formula for a cross product. We know that P Q Cross PR is equal to well, what's right in terms of you and be actually, for simplicity's sake, you cross V is equal to you too. VI three minus you. Three v two comma you three V one minus. You won B three and you one V two minus you, too V one. So we're gonna use the same formula to find the cross product. Except in this case, the victory you is going to be p. Q and Director V is going to be PR. So let's do the calculation Now, Peak, you cross PR is equal to you two times V three. So that's gonna be negative two times one, which is just negative, too. Minus you three times V two. So negative, too. Times three. That's just negative. Six. Next part you three times V one. So negative. Two times one is just negative, too. Minus you one times V three. So that's a negative. Two times one giving us negative, too. And third part you one times be too negative. Two times three is just negative. Six. Minus YouTube Times V one. That's negative. Two times one. Giving us a negative, too. Now, when we solve this, we get our final answer as negative. Two plus six is four negative two plus 20 and the negative six plus two is negative for. Therefore, our final answer is negative or four zero negative for.

Okay, so they want to find a product dot product of the following. That's the product of horizontal cavorted, sort of seven times for plus the product of our vertical components. So that's negative. Two times 14. So this is equal to 28 minus 28 which is equal to drop.

In this problem. We are given three points and told that a plane passes through those points. Then we were asked to determine a factor that is perpendicular to that plane. And so the way we're going to do this is to find the vector. We're going to take the vector p Q and across it with the vector PR. So now how do we find P Q and P. R is we take point Q. And subtract point key in order to get P. Q. Because Q is the endpoint and P is the starting point. So we're going to take 12 negative one and subtract 010 And when we do that, we get 11 negative one. Okay, so that is equal to the Vector p Q. And now he wants to find the doctor PR. And so to do that, we're gonna take the point are and subtract the point p. So when we do that, we get negative, too. 10 minus 010 And when we do this, we get negative 200 And now we're going to cross these two doctors once we turn them into doctors. And by turning them into vectors. They say the same, essentially, so the doctors are going to be 11 negative one, and the second back to is going to be negative, too. 00 Wonderful. And so now we're going to cross them by first reading of Matrix with I, J and K in the three stroke. The components of the Vector P Q. In the second row, which are 11 negative one and the components of the vector p R in the third row, which are negative 200 and now state the cross product. Get the components of I. We're going to cross that. The first column Take the determinant of this two by two matrix, which is one time zero minus negative one times. Here, let's see remind Syria that's zero to get the components of J, which we always subtract. Chase component going to cross at the second column and take the determinants of this to bow to Matrix, which is going to be 10 minus negative. One times negative two is going to be zero minus two. That's negative, too, and the negatives they're going to cancel in Korea positive then to get K's component. We're going to cross out the third column and take the determinant of this two by two matrix. It's going to be one time zero minus one. Have negative two. That's going to be zero minus native tours here, opposed to, which is just too. And we have found our doctor is going to be zero to to.


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