chrisplusian
  • chrisplusian
Someone please???....Find an equation of the plane. The plane that passes through the point (3, 4, 5) and contains the line x = 5t, y = 3 + t, z = 4 − t
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
chrisplusian
  • chrisplusian
@zepdrix ?
chrisplusian
  • chrisplusian
@thomaster ? @ganeshie8 ? @.Sam. ?
chrisplusian
  • chrisplusian
I have been trying to get help with this one all day

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chrisplusian
  • chrisplusian
@Jhannybean ?
chrisplusian
  • chrisplusian
Sorry I am stumped and am just reaching out for help
Jhannybean
  • Jhannybean
You know whats funny,i'm stuck on a similar type of problem.
chrisplusian
  • chrisplusian
What is yours? I just don't get this because it seems parametricsed
Jhannybean
  • Jhannybean
Youre going to need to derive the vector from the parameters youve found.
Jhannybean
  • Jhannybean
Do you know how to do that?
chrisplusian
  • chrisplusian
I believe I can derive a vector representation of a line but don't I need three points in order to find a "normal" vector? I thought to construct a plane you had to have three points or one point and a normal vector
chrisplusian
  • chrisplusian
|dw:1378694687919:dw|
Jhannybean
  • Jhannybean
you need your \(\large P_0\) (original point i'd like to say) and a normal vector \(\large \vec n = \)
Jhannybean
  • Jhannybean
youve got \[\large x=5t\]\[\large y=3+t\]\[\large z= 4-t\] and you know \[\large P_0= (x_0,y_0,z_0)\] and you know \[\large x=x_0+at \ , \ y_0=y_0+bt \ , \ z=z_0 =ct \]
chrisplusian
  • chrisplusian
If I know three points I can use the cross product of the two lines generated to find the orthogonal (normal) vector or if I had one point and the normal I could do it but I don't know how to use a parametric equation to do that. Would I just arbitrarily pick values of "t"?
Jhannybean
  • Jhannybean
Follow with me here.
chrisplusian
  • chrisplusian
ok I am with you
Jhannybean
  • Jhannybean
We have 1 point. \(\large P_0 = (3,4,5)\). we could find another point from out parametric equations using the formula \(\large P_0=(x_0,y_0,z_0)\)Do you see it?
chrisplusian
  • chrisplusian
am not following
Jhannybean
  • Jhannybean
Basically, our parametric equation will give us a directional vector, and a point.
chrisplusian
  • chrisplusian
And with that you can construct a line right? But how does that determine the equation of the plane?
Jhannybean
  • Jhannybean
Do you remember the formula for the equation of the plane?
chrisplusian
  • chrisplusian
\[ax+by+cz+d+0\] where is the normal vector?
chrisplusian
  • chrisplusian
or \[a(x-x _{0})+b(y-y _{0})+c(z-z _{0})=0\]
Jhannybean
  • Jhannybean
ax +by +cz +d = 0
chrisplusian
  • chrisplusian
sorry yes should have been =0 not +0
chrisplusian
  • chrisplusian
without a normal(perpendicular) vector that doesn't help does it?
Jhannybean
  • Jhannybean
we need to find a vector, our parametric equations can help us with that. \[\large x=\color{red}5t\]\[\large y=3 +\color{red}{1}t\]\[\large z=4-\color{red}1t\]\[\large \vec v = <\color{red}{a,b,c}>\]
chrisplusian
  • chrisplusian
ok
Jhannybean
  • Jhannybean
do you see how i get the vector
chrisplusian
  • chrisplusian
yes I am comfortable with that
Jhannybean
  • Jhannybean
Alright. We've got 1 vector, and 1 point now. We need 2 points to create another vector, and 2 vectors (cross product'd) to create a normal vector
chrisplusian
  • chrisplusian
agreed
Jhannybean
  • Jhannybean
Now using the parameters, we can find ourselves the SECOND point. \[\large x=5t+\color{red}0\]\[\large y= \color{red}{3}+t\]\[\large z=\color{red}4-t\] \[\large P=(\color{red}{0,3,4})\]
chrisplusian
  • chrisplusian
ok
Jhannybean
  • Jhannybean
Now we've got 2 points, and 1 vector.
Jhannybean
  • Jhannybean
Btw, do you have the answer in the back of your book? o-o
Jhannybean
  • Jhannybean
I just wanted to know if i was doing it right.
anonymous
  • anonymous
Ok let's first convert the parametric form of the line to vector form: [x,y,z] = [0,3,4] + t[5,1,-1] We note that the vector [5, 1, -1] is the direction vector of the line. We also note that this line lies on the plane and the equation of this plane will be in the form: Ax + By + Cz + D = 0 We also note that the vector [A,B,C] is a normal vector to the plane. We know that the vector [5, 1, -1] and so it's a direction vector of the plane. We also know that the plane goes through the points (3,4,5) and (0,3,4). Let's subtract these vectors to get another direction vector: [3,4,5] - [0,3,4] = [3, 1, 1] So now we have to two direction vectors for the plane which are [3,1,1] and [5, 1, -1] and they are not scalar multiples of each other. Now to find the normal vector [A,B,C], we must find a vector perpendicular to both direction vectors. To do this we can do two things: use cross product or dot product. I will use the dot product to keep things simple. Now since the vector [A,B,C] is orthogonal to both direction vectors then the dot product of [A,B,C] with each should be 0: [A, B, C] · [5, 1, -1] = 0 => 5A + B - C = 0 [A, B, C] · [3, 1, 1] = 0 => 3A + B + C = 0 Let's equate the two: 5A + B - C = 3A + B + C => 2A - 2C = 0 => A - C = 0 => A = C. Let's just choose 2 values of A and C that satisfy the equation. I choose A = 2 then C =2 as well. Plugging these values in to the original equations we get that B = -8. Then the normal vector [A,B,C] = [2, -8, 2]. Now let's substitute these values in to the equation of the plane to obtain: 2x - 8y + 2z + D = 0 We are given that the point (3, 4, 5) lies on the plane. So plug these values in to the equation and solve for 'D': 2(3) - 8(4) + 2(5) + D = 0 => D = 16 Therefore the equation of the plane is: 2x - 8y + 2z + 16 = 0 And we are done. @jhannybean @chrisplusian
anonymous
  • anonymous
You can also factor out the '2' from the equation of the plane to obtain: x - 4y + z + 8 = 0
chrisplusian
  • chrisplusian
Sorry I fell asleep last night. Thank you @genius12 and @Jhannybean

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