5

In each part, find two unit vectors in 2 -space that satisfy thestated condition.(a) Parallel to the line $y=3 x+2$(b) Parallel to the line $x+y=4$(c) Perpendicular...

Question

In each part, find two unit vectors in 2 -space that satisfy thestated condition.(a) Parallel to the line $y=3 x+2$(b) Parallel to the line $x+y=4$(c) Perpendicular to the line $y=-5 x+1$

In each part, find two unit vectors in 2 -space that satisfy the stated condition. (a) Parallel to the line $y=3 x+2$ (b) Parallel to the line $x+y=4$ (c) Perpendicular to the line $y=-5 x+1$



Answers

In each part, find two unit vectors in 2 -space that satisfy the stated condition. (a) Parallel to the line $y=3 x+2$ (b) Parallel to the line $x+y=4$ (c) Perpendicular to the line $y=-5 x+1$

Russian here gives us two equations. Why is equaled to expose three and two Y minus four X minus five is equal to zero, and it wants us determine whether the lines are parallel perpendicular. Neither So. First of all, let's rearrange the second equation so we can solve it in terms of why. So it's the same. So two y is equal to four X plus five. Or why is equal to two X plus 2.5 or five over to and from here, as we know, it's the point slope intercept form, which takes place using the equation Y is equal to X plus C. And as we can see, the two cold fish is the same, which denotes the slope. Um, we can say that these lines are therefore parallel.

The question here gives us two equations negative three X plus four winds before and four x plus three wide. Five wants us to determine whether or not the lines are parallel perpendicular. Earn either. So, first of all, let's solve both of these equations in terms of its wide with respect to its Bly. So let's do the 1st 1 first. So for y is equal to three x plus four. Therefore, why is equal to three over four X plus one? And in this second part three y is equal to negative four X plus five where why isn't the negative for over three X plus five over three. So as both of these take place in the form of the slope intercept form, which is I'm X plus C or the M tells us the slope, we can see that both these slopes are negative. Reciprocal sze of each other, therefore weaken state that the equations of the two lines that were given are perpendicular to each other. To Laura Rather Yeah,

So we have these two lines. We have negative two X plus y is equal to three, and we also have three acts plus three halves. Why is equal to five and we want to see? Are these parallel, perpendicular or neither? So that's we can do different forms. But you know, I like to put things into slope intercept form into that Y equals MX plus B. I just like that form because we know if the slopes are the same, then we will have parallel. We know if the slopes are opposite and reciprocal, then we'll have perpendicular. And if either of those happens, then we know that it's a neither. So this equation. Let's add the two extra both sides and we have y equals two X plus three and this equation putting it into slope intercept Farm. I'll just do it underneath, will subtract three x from both sides and then I'm gonna multiply everything through by two thirds and I'll distribute that two thirds to each of them over the two thirds over here. And so this becomes just good old why? And here we have that three reduces that three, so we get negative two acts plus, and then this will be over one. So that's gonna be 13th. So now we have them in the forms where we can compare, and I can see that they do not have the same slope. Therefore, we know they're not parallel question is, are the perpendicular well, they're opposite, but they're not reciprocal to be perpendicular to this one. I would need a slope up negative and then one half to be perpendicular to this line. I would have to have the other line have a slope up positive one half and that's not the case, so we have the condition of neither.

This question gives us two vectors and asks us to find a vector that's perpendicular to both of them. We have a special tool for this, and that's gonna be the cross product. We know that when we have two vectors a cross be that's always gonna be perpendicular to a and perpendicular to be. This was shown in your book by calculating a cross be dot product. A. The dot product, when two vectors are perpendicular, is always gonna be equal to zero. And we proved that no matter what A and B are a crust be dot a will equal zero. The same is true for be so a Crosby's perpendicular both two A and two B So let's jump into it the vectors that were given our A, which is equal to three j plus five k. Now, this is a linear combination of are unit vectors, which we can pretty easily turn into Just one factor. It's gonna be 035 The second vector that were given is B, which is equal to negative. I plus tu que again. That's gonna be negative. 10 and two. So let's start doing this. We know that to take the cross product. We're going to be, um, finding the determined of a three by three matrix whose first row, our unit vectors I j k are third and fourth or second and third row will be our vectors. So 035 and negative 102 So let's start. We know that taking the determined of a three by three matrix we're gonna do co factor decomposition. And I'm gonna use the first row because it's gonna be far easier than anything else. So what we'll have is the determinant of minor 11 which is just the vector or the Matrix without wrote one and column one 305 to multiple that might I that we're going to subtract, Remember, we're subtracting because of where J is in the Matrix because Element 12 and wanted to add up to an odd number. We're gonna have to subtract it. So we're subtracting minor 12 which is zero negative. One, 52 times j. And then we're adding again or adding minor 13 which is zero negative. 130 times k R unit Vector K. So now we can do this two by two Major sees taking the determiner of them is a breeze where six of minus zero, which is six. I we're gonna have zero minus negative five, which is positive. Five. Remember, we're subtracting it, so it's gonna be minus five J and then we're going to, uh, zero minus negative three, which is just gonna be positive. Three. So plus three. Okay. And this is our cross product. This is our resulting vector from his to, uh, original factors. So the vector six i minus five j plus three k, or in other words, six negative +53 is a vector that's gonna be perpendicular to both a and B. This isn't answered a part of the question. So if we have better a here and vector be here, our vector is something like this. Is it a cross be and we can see that it is at a right angle with a end with B. But hold on. We also want to find a unit vector that is perpendicular to both of these. Well, how can we do that? We're gonna have a special tool, which is not really a secret. Is just scaling this cross product no matter what the length of this cross product is, it's always going to be perpendicular. So we could double its length, give half its length. We could divide its length by pie, no matter what the vectors, and be perpendicular to A and B. So we want a unified A unit vector has a magnitude of one. What we want to do is to get a magnitude of one. We just want to divide or scale this vector down by its own magnitude. The magnitude right now is this sum of the square root of the some of each component squared. So six squared plus negative five squared plus three squared. When we square all of those add them up. What we get is the square root of 70. So right now this factor has elect of route 70 we wanted to have a length of one. So all we have to do is divide it by its length. So our unit vector is gonna be one over squared of 70 times six negative, five three. So this is a unit vector that is perpendicular to a and B and that's your final answer


Similar Solved Questions

5 answers
Are the columns of the matrix 4 linearly dependent; or linearly independent? Give reasons~10
Are the columns of the matrix 4 linearly dependent; or linearly independent? Give reasons ~10...
5 answers
Take the inverse Laplace transform of the following functions 4s2 + 58 - 3 s ~ 1)(s + 2)(s + 1) 8 +4 2s + 2)(s + 2) 3s2 -1)?(s + 1)
Take the inverse Laplace transform of the following functions 4s2 + 58 - 3 s ~ 1)(s + 2)(s + 1) 8 +4 2s + 2)(s + 2) 3s2 -1)?(s + 1)...
5 answers
Illustrate the result that using the first-order total differential leads t0 under- estimates of the change in a function value for the function+x3_GtjkUse the initial point _ 1 = (1,.2) and changes in the x, values of dri = 3 and dx = |
Illustrate the result that using the first-order total differential leads t0 under- estimates of the change in a function value for the function +x3_ Gtjk Use the initial point _ 1 = (1,.2) and changes in the x, values of dri = 3 and dx = |...
5 answers
Part 3 Question 1 If the original 2D figure is dilated with center at the otigin $0 that tle base along the x-axis Was doubled in size; while its height along the yaxis Was teduced to % of its otiginal length; what would the coorditates of the new vettices be? Label thc new graph Graph of Original 2D) Figure Graph of New 2D Figurc
Part 3 Question 1 If the original 2D figure is dilated with center at the otigin $0 that tle base along the x-axis Was doubled in size; while its height along the yaxis Was teduced to % of its otiginal length; what would the coorditates of the new vettices be? Label thc new graph Graph of Original ...
5 answers
I00Quesdon 3 / 10 OjXGhe capacilor showa in figure initially uncharged If the switch $ closed at (=0 Find the time (in second) when potential difference equal to the potential deference across the resistance Given that R-30 KQ cuoss the capacitor is00.7 2014 34021 02.8 5.035
I00 Quesdon 3 / 10 OjXG he capacilor showa in figure initially uncharged If the switch $ closed at (=0 Find the time (in second) when potential difference equal to the potential deference across the resistance Given that R-30 KQ cuoss the capacitor is 00.7 2014 34021 02.8 5.035...
5 answers
Estimate the pKa values for the functional group classes represented by the given molecules.OHOHAnswer Banking InccontactushelpNH;
Estimate the pKa values for the functional group classes represented by the given molecules. OH OH Answer Bank ing Inc contactus help NH;...
5 answers
Find the equation of the circle.Center (-1,-3)$;$ passes through (-4,-2).
Find the equation of the circle. Center (-1,-3)$;$ passes through (-4,-2)....
1 answers
In Exercises 112–113, show that $$ 1+2+3+\cdots+n=\frac{n(n+1)}{2} $$ is true for the given value of n. $n=5 :$ Show that $$1+2+3+4+5=\frac{5(5+1)}{2}$$
In Exercises 112–113, show that $$ 1+2+3+\cdots+n=\frac{n(n+1)}{2} $$ is true for the given value of n. $n=5 :$ Show that $$1+2+3+4+5=\frac{5(5+1)}{2}$$...
5 answers
Fvalate the limit (voli mav nofase the granh)" lim _ (cosx)tanx
Fvalate the limit (voli mav nofase the granh)" lim _ (cosx)tanx...
5 answers
Graphing calculator TecommendedUse the graph of the function f t0 state the value each limit; if it exists_ (Ifan Jnawer doee not exist; enter DNE: )Kx)Kx)DNE
graphing calculator Tecommended Use the graph of the function f t0 state the value each limit; if it exists_ (Ifan Jnawer doee not exist; enter DNE: ) Kx) Kx) DNE...
5 answers
Divide: Write the quotient in simplest form:5 -3 Select the correct choice below and, if necessary; fill in the answer box to complete your choice_5+3 (Type an integer or simplified fraction ) The quotient is undefined.
Divide: Write the quotient in simplest form: 5 -3 Select the correct choice below and, if necessary; fill in the answer box to complete your choice_ 5+3 (Type an integer or simplified fraction ) The quotient is undefined....
5 answers
A thin vertical column of height fixed at both ends_ supporting mbes? P experiences force which tries to deflect it from the vertical If y(z) the sideways deflection_ height above the ground_ then satisfies the differential equationdrz myY(o) = 0, y (0) = 0.where € is constant depending the thickness and materials of the column_ For this problem assume that € and find the smallest mass for which non-zero solution exists _
A thin vertical column of height fixed at both ends_ supporting mbes? P experiences force which tries to deflect it from the vertical If y(z) the sideways deflection_ height above the ground_ then satisfies the differential equation drz my Y(o) = 0, y (0) = 0. where € is constant depending the...

-- 0.020672--