let-mathbfv_1leftbeginarrayr1-0-2endarrayright-mathbfv_2leftbeginarrayr-3-1-8endarrayright-and-mathb

Question

Let $\mathbf{v}_{1}=\left[\begin{array}{r}{1} \ {0} \ {-2}\end{array}ight], \mathbf{v}_{2}=\left[\begin{array}{r}{-3} \ {1} \ {8}\end{array}ight],$ and $\mathbf{y}=\left[\begin{array}{r}{h} \ {-5} \ {-3}\end{array}ight] .$ For what
value(s) of $h$ is $\mathbf{y}$ in the plane generated by $\mathbf{v}_{1}$ and $\mathbf{v}_{2} ?$

Runpeng Li · Accepted Answer

Okay, so for a problem 18 we wouldn&#x27;t want to find the value of age. So that, uh, Victor why he&#x27;s saying that he&#x27;s on the plane generated to buy of You want to be too. So, by the giving information we know first that next next to Sorry, I&#x27;ll just first right down. Do you want me to? So you wanna be too? Will be 10 connected to and neck of 318 And Victor, why will be age next five and collective three. So that is actually to consider this Al Commended Matrix age Negative five. Elective three. We want this mission this matrix to be consistent. That is the say, a matrix v spend by spend by V one and V two. So we&#x27;re considering the system. He actually was. Why? Okay, So we need to do what we need to do here is to performed the cash indignation. So it&#x27;s not a complicated matri. So let&#x27;s do this by hand. So way first used just 1/3 row plus twice up the first rule. So row three, us twice off row one. So we can cancel the first entry on the third row. So He&#x27;s a general and connective six here and plus eight. So we&#x27;ll be inactive. Started positive, too. And for the constant term, we have connective three plus two age. Okay. And I&#x27;ll keep the kitty out of bettors. One negative three and a JJ. Okay, so the next we want to cancel the third the second entry off on 1/3 row. So to do that, we we use Ah, the role minus twice off the second roll. So, bro. Three minus twice off road, too. So it will be one first. Right down. Ian changed those, and we have 01 and negative five. So for the last roll, there are zero to minus two times one. So that zero and negative three plus two age miners two times negative. Five. So? So it&#x27;s ah, um, right down. So elective three lasts to age minus two times. Row two will be negative. Five. So here it will be naked. Three plus two age plus 10. And this will be seven thus to age. Okay, so too to make they&#x27;re all commended me may trace. Um, consistent. We need to make the last entry on the right hand side to be zero. So we require, in this case, age to be inactive. Seven have so that day sentry will be zero than the system can be solved. Otherwise the system cannot be solved.

Sean Thrasher · Answer

Okay. So for what value of age gives us, that bee is in the span of a one and a two. Now, anyone in a two are vectors that sit in our three, right? So they may Look, you know, something like this, for example, right? And they span a plane. Okay, Well, you could see clearly that there linearly independent. And so they spawn Spy on the plane. Meaning that if you take all the points in our three that could be reached by some than your combination of a one and a two, you take all possible in your combinations scale and added up you get a plane, okay? And then we take be whose x and y components are fixed. Right? We&#x27;re trying to find the set component Z component such that the tip lies exactly on the plane. Okay. And so the question is what X value H gives us that be lies on the plane that spanned by a one and a two. Now, what does it mean for B to be in the span of a one and a two? It means that B is x times a one plus x times a two for some X one and X two in our right. So x one the next to exist, you can write. You can write down Be like this. Okay, now let&#x27;s convert this into a system. So if we plug in what a one and a two actually are. So we get that X one times one for negative too, plus X two times minus two minus 37 is equal to 41 h, right. And so we get this system of equations right here. So, like many things in the near algebra, like most things in the near algebra, this problem reduces too. The question of solving a system of linear equations. Okay, so we want to find what age is. Okay. And to do that, we need to find what ex one next to is first. So clearly we can use the 1st 2 equations to get that right. Two equations into unknowns. From these two equations, we get X one, the next to our and once we get what ex one next to our then we can use the third equation to find H. So I&#x27;ve written down the 1st 2 equations. Let me remind you, it&#x27;s x one minus two times X two is equal to four and four times X one minus three X two is equal to one. I&#x27;ve written that down as on augmented matrix here. Okay, so now what does this boil down to? It boils down to row reducing. Okay, so let&#x27;s roll. Reduce. So the first thing you can see here is that you want to get rid of the lower bottom. So left lower entry right here. And so you have to do Row one. Row two goes to road to minus four times. Row one. Okay, So wrote to goes to road to minus four times were one. So row one is untouched on road to is well, four minus four, minus three plus eight and then one minus six. Seeing yet, which is minus 15. And so clearly we can divide, you know, the second row by five by easily. So row five goes to one over five times. So row three goes to won over five times roll three. And so when we do that, we get this. Okay, so we get one minus 240 minus three. And finally I want to get rid of this entry right here to get into produce Roche long form and so wrong. One goes to throw one plus two times 02 So Row one changes this time on road to is left untouched. And so we get that X one is minus to an X two is minus three. That&#x27;s our solution. So H is equal to well, using the third equation. It&#x27;s this and then plugging in what x one and X two were. Get that. And so it&#x27;s four minus 21 so you get minus 17 and we&#x27;re done.

Linh Vu · Answer

the question of the plan and sustaining contends that three point we condemn is I, b and C here. So the 1st 1 will be 11 The second point is 101 on the last one will be 11 zero now way. I would use the point in here and you find a normal vector. First, I will find a bit I&#x27;d be and we go do I would tend the Beamon&#x27;s ice again the one month 10 I&#x27;m so we will find a better BC. It would ICO Geo is 01 minus one. And from here we can find a normal vector. It will be equal to that. I would be We can split the better BC and when we do so we should again the cross border of them were equal to one. Here, this one will be one, and this one will be one. So from here we can find a question in the plan. It will be one times x minus. There are plus what? Plus one times y minus one last one times Z minus one. But it got you there. 00 it was something again the x plus Why, plus the ICO to the Jew

Debra Leonard · Answer

in this problem were given a formula and values to substitute into the formula to find the missing piece. In the formula, there are 1234 actual variables. Capital A is a variable lower case. A is very well, B is avail variable and H is variable. And of those four, they&#x27;re giving you three and asking you to find the fourth. So we&#x27;re going to simply substitute in the values. So we&#x27;re going to take, for example, we&#x27;re going to take a and put it where a is every way to take B and put it where B is and we&#x27;re going to take a church and substituted in for ages. So we&#x27;re going to say that a is equal to 10 plus 16 time seven, divided by two and then order of operations have us take 10 plus 16 to get 26 times seven and divided by two. What I would do at this point is I would take 26 to and do that division, which would give me 13. So then it&#x27;s come over here. The area is going to equal 13 times seven, so the area is going to equal 91. It is a straight substitution problem

Urvashi Arora · Answer

Hello. The question is taken from victor&#x27;s and geometry of the space. And The question is find the equation of the plane and the plane passing to the 3.2123 -86 and -2 -31. So let me right equation or plain passing to three points 212 3 -86 and 30s -2 -31 can be read Ernest. So let me write it days X -2 x -1 that minus two. And here three minus two minus eight minus 16 minus two and thirties minus two minus two minus three minus one and 1 -2 is equal to zero. So let me so always we get X -2 into they send this so we get- dining to minus when it&#x27;s nine minus four into four minus 16. So here comes the plus sign plus 16 and the second one is minus by minus one into K -2 into 1 -2. So that will be -1 minus off -2 -2. It&#x27;s -4 -4 in two plus 4 plus minus four Plus 4 16. So that will be plus 16. So plus that minus to win too -3 -1. That will be -4 into 1 -4 -9 in two -4. So that will be plus 36. Sofia we get -36 is equal to zero for those solving it. We get 25 x -50 minus y minus went into 15 Plus Zed -2 into -40 Is equal to zero. And further we get 25 weeks minus 50 -15. y plus 15 minus 40. 0 Plus 80 is equal to zero. And we get the value 25 weeks, okay -15 Y -4, visit 80 and 15 is 95 95 -50 45, so that will be 45 which is equal to zero. Taking five women, we get five weeks -3 Y ages eight plus nine is equal to zero. So the required answer for this question is five x minus three by- Exit is equal to -9 which is the required and some of discussion. Hope this clears your doubt and thank you.

Problem

Give a geometric description of $\operatorname{Sp…

Problem 18 Hard Difficulty

Answer

$h = - 3.5$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$h = - 3.5$

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Maria G.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications