5

Point) Supposefkz) = 2+1and9(m) = I+1.Tnan A (J + 9)(-) 0. (J 9)(2) = (fg)z) = D. (f/9)r) E(f 0 9)(z) E(go f)r) =and...

Question

Point) Supposefkz) = 2+1and9(m) = I+1.Tnan A (J + 9)(-) 0. (J 9)(2) = (fg)z) = D. (f/9)r) E(f 0 9)(z) E(go f)r) =and

point) Suppose fkz) = 2+1 and 9(m) = I+1. Tnan A (J + 9)(-) 0. (J 9)(2) = (fg)z) = D. (f/9)r) E(f 0 9)(z) E(go f)r) = and



Answers

Refer to the figure.
If $J M=3$ and $M K=9,$ find $L J$

We want to express the vector going from a to B in the form off this I j k notation. So let's first go ahead and turn these points into vectors using that I j k notation and then we can go from there. So remember, this is going to be the Eid component, the jade component and Kate component, respectively. And each of these are just going to be the coefficients off i, j and K. And then we add them all up. So is going to be negative. Seven I plus a negative eight j. So that's just minus a J. And then plus one K which we would just right is K now over here for B. We're gonna do the same thing. So again I, j and K So this is going to be negative. 10 I plus eight j plus one. Okay, so we have our two points now to figure out the factor going from a to B. We look right above it. It says we can use this formula words just the point. We want to land on minus the point we're starting from, So this is going to be being minus a so we could go ahead and plug goes in. So be is negative. 10 I plus eight j plus K. And they were going to subtract a which is negative seven I minus eight J plus K And just like when we had polynomial expressions we could add are like terms. So we're gonna go ahead and do that here. So we have negative 10 I then minus a negative seven I so that's gonna be like adding it. So we should have minus three i for our 1st 1 and then we're gonna have a J minus a negative h a. So we're gonna add those as well, because the negatives cancels. That's gonna be plus 16 j and then we're gonna do K minus K, which is going to give us zero k. So you just go ahead and erase that her. So our vector that goes from point A to point B is going to be negative. Three i plus 16 j

For this exercise. Let's start by considering this expression, which represents and find combination off the vector is 51 52 53. Let's set this equal to a plus TV, which is a point on the Given Ray on. We can rewrite this expression as C two times 50 to minus 51 plus C three times 33 minus V one plus T times minus V equals two a minus fever. So with this, with factors that were given 50 to minus V one is the victory. Six year old, one feet three minus 51 is a vector 264 uh, then crusty and minus 1.4 minus 1.5 on day 3.1 on a minus. 51 is the victory minus one minus three and 15. So what we want to do is we want Thio find values for C two C three and t that satisfy the equation and that we do that by solving these matrix equation. Andi, once we solve, we can do that by rendition. Ornamented my tricks. For example, we get C two equals 0.6 c three equals 0.5 on t equals four on With these two values, we can see that one minus six to minus E three is equal to minus 0.1. So then, when we subsided in our original expression, we get minus 3.131 plus to a 0.6 ft. Two plus 0.5 ft three is equal to a plus four B On this is a vector 5.6 six on minus 3.4. This vector is a point of intersection between the ray on. They are fine or off. The one V two V three now looked at the first coefficient is negative on. This is a sign that the point of intersection it's outside the triangle region, given five, the one doing feet three.

Okay, We want to write the vector going from a to B in this I j k notation. So let's first go ahead and rewrite our points A and B in terms of a vector with his i j k notation. So for a remember, I, J and K are going to be the X y and Z component, respectively. And we're just going to essentially make each of these components the coefficient of I J K and then Adam Allah. So there's gonna be one times I So that's just gonna be I plus zero times j Well, that's just going to go away and then plus three, case this actually, just go ahead and cross that out now for be so again I, j and K each of them are going to be the coefficients of that component. So it's going to be negative. I plus or times J plus five times k. So now we have our points wrote, hasn vectors. Now what we can do is use the formula to tell us that goes from a to B. So this is just going to be being minus a So let's go out and plug that and so be, is the vector negative? I plus four J plus five k and then we're going to subtract this hack. So But it so x is supposed to be Hi, plus three K. And now we can go ahead and add these just like we would Any polynomial expression Combining are like terms. So we're gonna have negative I minus. I says, gonna give us negative to I They were gonna four j minus. Well, there's no forge a term or there's no J term over here, so it's just gonna be plus or J and I'm gonna have five k minus three Kate, which is going to give us plus two k. So this will be our vector that goes from A to B and it's in the I J K notation.

So if we're given that the vector A to B is I plus for J minus two K and the point B is +513 We want to find what a it's So first, let's go ahead and rewrite a Be using the bracket notation just because I prefer that notation more. Um, so I was coefficient is tthe e x component. Jay's coefficient is the why component and Kay's coefficient is the Z component. Negative too, are not negatively but negative too. So we have that. Now let's think about what it means to be the vector A B So if we start from the plane A and we add the vector Abie to it, we should land on be so since we want to find a is, What we're gonna do is subtract a B over. I'm gonna get a is equal to B minus a B. What? So now we can't go ahead and plug these bodies and so I'm gonna plug in B as a director. So five one three minus and a and A B, we said was one for minus two. Now, remember, when we subtract two vectors, we subtract them component wise there's gonna be five minus one, which is going to be or and then we're gonna do one minus or just going to be negative three and then three minus negative, too, which is going to be five. So, um, since a really is in a Vectra, let's just go ahead and write it using parentheses instead of these brackets here, so a is going to be the point or minus three.


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