Question

Vectors:

Determine the Cartesian equation of the the plane that is parallel to the line with equation x = - 2y = 3z, and that contains the line of intersection of the planes with equations x - y + z =1 and 2y - z =0.

correct answer is : 8x+14y -3z -8=0
how did they get this?

Answer 1

ganeshie8, what do you think? I got the intersection line between 2 given planes is 2x -3y-3z=0

Answer 2

@Data_LG2 x =-2y=3z is the symmetric equation of the line?

Answer 3

yes, i guess so..
how will i find the normal vector of x =-2y=3z?

Answer 4

we need direction vector, not normal one for the line

Answer 5

I am sorry, I have to go . Again, I am sorry.

Answer 6

oh yes.. sorry lol

Answer 7

so how will i find the direction vector of that line?

Answer 8

This is probably not the most efficient method, but it works ;)

You probably know that you can specify a plane by a position vector that determines an arbitrary point on the plane and two direction vectors than span the plane, so to speak. So what use is that here? Well, we get the position vector and one of the direction vectors from the line of intersection mentioned in the problem, since that line lies within the plane we are about to determine.

And the second direction vector? That comes from the direction of the line that is to be parallel to our plane. But what if that second direction vector happens to be parallel to the first one (we would not be able to span a plane in that case)? Well, then there wouldn't be a unique plane in the first place and the problem would not make much sense, so we can safely assume that they are not parallel (but we could check for that, if we wanted).

Anyway, let's first determine the line of intersection. One possibility is to take both equations, treat them as a system of linear equations and solve that (partially) with any method we like, such as Gaussian elimination, substitution or whatever.

x - y + z = 1 [A]
2y - z = 0 [B]
-----------------------
x + y = 1 [A+B]

Now we can choose 2 arbitrary pairs of x and y that satisfy the last equation [A+B] and use them to find the respective values of z and so acquire 2 points that lie on the line of intersection:

x = 0, y = 1

x - y + z = 1 or 2y - z = 0
0 - 1 + z = 1 or 2*1 - z = 0
z = 2 which equals z = 2

So our first point is (0, 1, 2).

x = 1, y = 0

x - y + z = 1 or 2y - z = 0
1 - 0 + z = 1 or 2*0 - z = 0
z = 0 which equals z = 0

So our second point is (1, 0, 0).

We can just take one of the points as position vector for our plane. Let's choose (0, 1, 2).
And by subtracting one point from the other (doesn't matter which way around) we determine our first direction vector: (1, 0, 0) - (0, 1, 2) = (1, -1, -2)

For the second direction vector we can either convert the given equation for the line into a parametrized equation or again simply find 2 arbitrary points on that line by choosing two values for one of the coordinates and calculating the respective other two:

z = 0

x = -2y = 3z
x = -2y = 3*0
x = 0 as well as y = 0

Point #1: (0, 0, 0)

z = 1

x = -2y = 3z
x = -2y = 3*1
x = 3 as well as y = -1.5

Point #2: (3, -1.5, 1)

We don't need a position vector now, only a direction vector, so let's subtract immediately: (3, -1.5, 1) - (0, 0, 0) = (3, -1.5, 1) [do not forget to subtract - that one of our points was (0, 0, 0) is just a coincidence, it's not generally true]

So in summary, we have position vector (0, 1, 2) and direction vector (1, -1, -2) as well as (3, -1.5, 1).

The parametrized equation (using parameters r and s, but this is arbitrary) for our plane would be
(x, y, z) = (0, 1, 2) + r * (1, -1, -2) + s * (3, -1.5, 1)

If required, we can convert this plane into the equation without vectors. Either by using a system of linear equations and eliminating r as well as s or by using some vector manipulations. I prefer the latter method:

Let's first convert the plane in normal form by calculating a normal vector from the 2 direction vectors by doing a cross product:

(1, -1, -2) X (3, -1.5, 1) = (-4, -7, 1.5)

The position vector is simply copied over:

(-4, -7, 1.5) ∙ ( (x, y, z) - (0, 1, 2) ) = 0

Now simply expand:

-4 * (x - 0) - 7 * (y - 1) + 1.5 * (z - 2) = 0
-4x - 7y + 7 + 1.5z - 3 = 0
-4x - 7y + 1.5z + 4 = 0

That's it. It doesn't look like the given answer? Well, since it's an equation, we can multiply both sides with an arbitrary factor (apart from 0), such as -2:

-4x - 7y + 1.5z + 4 = 0 | * (-2)
8x + 14y - 3z - 8 = 0

Answer 9

i saw this: http://ca.answers.yahoo.com/question/index?qid=20110612141926AALYSCi
well, thanks anyways..

i'm still trying to understand his explanation though

Answer 10

@KingGeorge please, help. I forgot how to find the direction of the line x =-2y = 3z

Answer 11

@primeralph

Answer 12

@Loser66 Find 2 points P1 and P2, then find P2-P1.

Answer 13

I need explanation and particular method. It 's helpful than copy and paste

Answer 14

@TanteAnna method should work. In this case, notice that we have the point (0,0,0) on the line. Then we can take an arbitrary \(x\) value, and see what \(y\) and \(z\) values pop out. So for example, \(x=6\), \(y=-3\), and \(z=2\). would be one possible point. So the direction of the line is \[\begin{pmatrix}
6\\-3\\2
\end{pmatrix}\]and you can even say that the line is equal to\[t\begin{pmatrix}
6\\-3\\2
\end{pmatrix}\]

Answer 15

ok, so, it 's done, right? we have the direction of the plane, the intersection line of the given planes , take cross product of those lines to get normal vector of the required plane and put it in the formula. , right?

Answer 16

hmm..give me a sec.
i'll try to do it until the end

thanks !