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Find an eguntion %C the targcet line c thc givn curv? at the spiied point: (2) (2 _ %+20)7 82% (9-(,.2),Fic:|he vlitsi cf Ed Llut #ill mueke tle giv t fuzetic t eon...

Question

Find an eguntion %C the targcet line c thc givn curv? at the spiied point: (2) (2 _ %+20)7 82% (9-(,.2),Fic:|he vlitsi cf Ed Llut #ill mueke tle giv t fuzetic t eontinuaus ntcl :lffereuciabic everywher? ia ~6o, Co):0+%-2 # +dr + u,T <-1{(#)12-[

Find an eguntion %C the targcet line c thc givn curv? at the spiied point: (2) (2 _ %+20)7 82% (9-(,.2), Fic:|he vlitsi cf Ed Llut #ill mueke tle giv t fuzetic t eontinuaus ntcl :lffereuciabic everywher? ia ~6o, Co): 0+%-2 # +dr + u, T <-1 {(#) 12-[



Answers

Find a point $P$ on the line and a vector $\mathbf{v}$ parallel to the line by inspection. $$ \begin{array}{l}{\text { (a) } x \mathbf{i}+y \mathbf{j}=(2 \mathbf{i}-\mathbf{j})+t(4 \mathbf{i}-\mathbf{j})} \\ {\text { (b) }\langle x, y, z\rangle=\langle- 1,2,4\rangle+ t\langle 5,7,-8\rangle}\end{array} $$

In this problem for parts A and B. We want to find a point that's on the line and a non zero vector that's parallel to the line. And in order to do that, let's recall that if we have a line with a point P. It's called this peanut and a vector that's parallel to the line which is V. Well we can find the equation for this line by taking an arbitrary point on the line Q. And using the fact that the vector with initial point peanut and terminal point Q is parallel to the vector V. So what this means is that the vector from P to Q is a scalar multiple of the vector V. And we can rewrite this as vector with components X minus X. Not? Why minus? Why not? Z minus Xena is equal to T. And then for the vector V. We can use the components abc and we can simplify this again to get the following equation for a line. But why is this helpful? Well we can see that this factor with components X not? Why not see not? That corresponds to a point on the line, namely peanut. And this factor with components abc. Well that was a vector V which was parallel to the line. So with this information we can use that for parts A and B. By taking P A. Point on the line to be the point given by the coordinates 2 -1 for part a And -1- four. In part B. And next we have that the vector V that's parallel to the line in part A Is simply the vector with components for -1 and in part B. This is a vector with components 57 and minus eight. And that completes the problem.

In the question given the equation virtualized, the first line is 13 equals ju the 011 aan den plus the tee times the +112 And we have the art too. But here would ask her equal to the +203 plus as times the one for For now we want to find the intersection between them. So when did you get the I want Thio Put them equal to the R two s, so doesn't oblige them. We should have the system of questions. Here we have the zero plus the most echo into the two plus asked are then one plus the must equal to the zero plus for ask, then one plus two the most equal to the three plus four ask and then they try to serve the system in question. Now here to serve it, we take the first on the second and the third one. Here we do the subjection and we do so different guns zero then minus three equal to the minus three plus zero doesn't oblige. That team is equal to the three. When going to take three, you can block back to the first equation here, You get the answer. They were equal to the three minus two, equal to one. And now we need to make so that you go to certify the same problem intersection. So it will be good for the T equal to three. We will bring it to the R 13 that were equal to the 011 plus the three times +112 So we should get equal to the point. Here will be three and then one plus three. Could you far one plus six equal to seven. And now let's try to check if the ethical to one are too one, it will be good to the 203 plus the 144 So if we add them up with a skeptical to the 34 and seven so they much of each other here. Therefore, we have the problem. In the section will be this point here

Hello here. We're going to start program number seven here. The slope of the line is killed. Club off Group Point is given We just minus two the two points or zero comma de and my coma for here. We need to find T Can be found out by floor equals changing light by changing exp floor piece minus two. Witches equals four minus t. We were led by nine miners remain. So if we saw this bigger day equals 22. Thank you.

For this problem. We are asked to show that R. One and R. Two to find the same line where R one equals three negative 14 plus t times 8 12 negative six. And R two equals 11. 11 negative two plus two times 46 negative three were given a hint to show that our two passes through three negative 14. And that the direction vectors for R. one and R. two are parallel. So let's start with what the hint first indicates. Now let's see here. So You can see that if we want to pass through the .3 we need to take a negative value for T. Now 11 -4 is not going to equal three. In fact that will equal seven But 7 -4. Well give us three. So um or rather not seven yeah seven miles four. So if we take our starting point and subtract off to t. Then we'll get actually I'll write this out explicitly. We'd have that are two of negative two Would be 11. 11 negative too minus To. Actually I'll write this out explicitly. Plus negative eight negative 12 positive six. Which is then going to be the .3 -14. So we can see that we do pass through that point then to show that the direction vectors are parallel. Well the direction vector for our one which I'll say um just D. i. r. of R. one eight, 12 negative six. But we can reduce that If we divide by two we don't change the direction but we get 46 -3. And we can see that that equals the direction vector are too So since we can change the, since we have the direction vectors are the same or are parallel, and we do pass through the same points. That shows that yes, they are the same line.


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Wue;E " #ai3_07 7836
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