Two planes intersect in a straight line l,how to find the pt on it,(3j,2k)?

[inv]

[Solved]Two planes->x+2y-2z=2 & 2x-3y+6z=3 intersect in straight line l

Homework Statement

Hi.I've two planes' equation x+2y-2z=2 & 2x-3y+6z=3.The planes intersect in the straight line l. The question:Find a vector equation for the line l.

Homework Equations

a.n=0 if they're both perpendicular 2 ea other
r.n=a.n ,where n=perpendicular to both r and a
3 dimensional vector equation formula:{A position vector of line}+t{direction vector of line},where t is a variable.

The Attempt at a Solution

1(x)+2(y)-2(z)=0
1(2)+2(1)-2(2)=0
$$\left(\begin{array}{cc}2\\1\\2\end{array}\right)$$

x+2y-2z=2
(2)+2(1)-2(1)=2
$$\left(\begin{array}{cc}2\\1\\1\end{array}\right)$$

$$r_{1}$$=$$\left(\begin{array}{cc}2\\1\\1\end{array}\right)$$+t$$\left(\begin{array}{cc}2\\1\\2\end{array}\right)$$

2x-3y+6z=0
2(1)+3(8/3)+6(1)=0
3$$\left(\begin{array}{cc}1\\\frac{8}{3}\\1\end{array}\right)$$=$$\left(\begin{array}{cc}3\\8\\3\end{array}\right)$$

2x-3y+6z=3
2(3)-3(7)+6(3)=3
$$\left(\begin{array}{cc}3\\7\\3\end{array}\right)$$

$$r_{2}$$=$$\left(\begin{array}{cc}3\\7\\3\end{array}\right)$$+m$$\left(\begin{array}{cc}3\\8\\3\end{array}\right)$$

Putting $$r_{2}$$ in 1st plane cartesian equation getting:
(3+3m)+2(7+8m)-2(3+3m)=2
m=$$\left(\begin{array}{cc}\frac{-9}{13}\\\end{array}\right)$$

$$r_{2}$$=$$\left(\begin{array}{cc}3\\7\\3\end{array}\right)$$+m$$\left(\begin{array}{cc}3\\8\\3\end{array}\right)$$
=$$\left(\begin{array}{cc}3\\7\\3\end{array}\right)$$+$$\left(\begin{array}{cc}\frac{-9}{13}\\\end{array}\right)$$$$\left(\begin{array}{cc}3\\8\\3\end{array}\right)$$
=$$\left(\begin{array}{cc}\frac{12}{13}\\\frac{19}{13}\\\frac{12}{13}\end{array}\right)$$...$$\alpha$$

Putting $$r_{1}$$ into 2nd cartesian equation getting:
2(2+2t)-3(1+t)+6(1+2t)=3
t=$$\frac{-4}{13}$$

$$r_{1}$$=$$\left(\begin{array}{cc}2\\1\\1\end{array}\right)$$+$$\left(\begin{array}{cc}\frac{-4}{13}\\\end{array}\right)$$$$\left(\begin{array}{cc}2\\1\\2\end{array}\right)$$
=$$\left(\begin{array}{cc}\frac{18}{13}\\\frac{9}{13}\\\frac{5}{13}\end{array}\right)$$...$$\beta$$

$$\beta$$-$$\alpha$$=$$\left(\begin{array}{cc}\frac{18}{13}\\\frac{9}{13}\\\frac{5}{13}\end{array}\right)$$-$$\left(\begin{array}{cc}\frac{12}{13}\\\frac{19}{13}\\\frac{12}{13}\end{array}\right)$$=$$\frac{1}{13}$$$$\left(\begin{array}{cc}6\\-10\\-7\end{array}\right)$$

l's vector equation=$$\left(\begin{array}{cc}\frac{18}{13}\\\frac{9}{13}\\\frac{5}{13}\end{array}\right)$$+w$$\left(\begin{array}{cc}6\\-10\\-7\end{array}\right)$$


Anything wrong thus?

 

[HallsofIvy]

I don't understand what you are doing! You seem to be consistently confusing planes with lines.

3. The Attempt at a Solution
Direction vector for 1st plane:
1( x)+2(y )-2(z )=0
1(2)+2(1)-2(2)=0
(2i,j,2k)

Yes, 2i+ j+ 2k is one out of an infinite number of vectors lying in the first plane but knowing that doesn't help you. It is not the "direction vector" for the plane. Of course, it is also true that 1(3)+ 2(-1/2)- 2(1)= 0 so i- (1/2)j+ k is another such vector. You may be confusing this with the "direction vector" of a line: a vector pointing in the same direction as the line. The closest thing to the "direction vector" for a plane is its normal vector.

Pt for equation of 1st plane:

x+2y-2z=2
(2)+2(1)-2(1)=2
(2i,j,k)

Yes, 2i+ j+ k is the position vector of a point (2, 1, 1) which lies in the plane- but since you have no reason to think that it also lies in the second plane that doesn't help either.

Vector equation of 1st plane,r1=(2i,j,k)+t(2i,j,2k)

Absolutely not! A plane does NOT have a "vector equation" in one parameter- since a plane is two dimensional, any equation for a plane requires two parameters- since you can solve one equation in 3 variables for one of the variables in terms of the other two, the equations you are given are in "two parameters". What you have here is the vector equation for one of the infinite number of lines lying in the first plane.

You have two equations for 3 variables, x, y, and z. Go ahead and solve for two of them in terms of the third, then use that last variable as the parameter.

For example, so that you can do the problem you were given yourself, suppose the equations were x- y+ z= 0 and 2x+ y+ z= 2. Adding the two equations eliminates y: 3x+ 2z= 2. Then 3x= 2- 2z so x= 2/3- (2/3)z. Replacing x by that in the first equation, 2/3- (2/3)z- y+ z= 0 so y= 2/3- (2/3)z+ z= 2/3 + (1/3)z. Using z itself as parameter (but renaming it "t") we have x= 2/3- (2/3)t, y= 2/3+ (1/3)t, z= t as parametric equations for the line of intersection of the planes. Or, if you don't like fractions, take z= 3t: x= 2/3- t, y= 2/3+ t, z= 3t are the parametric equations. The vector equation would be, of course,
$$\vec{r}= (\frac{2}{3}- t)\vec{i}+ (\frac{2}{3}+ t)\vec{j}+ 3t\vec{k}$$.

By the way, write your vectors as (a, b, c) or ai+ bj+ ck. Do not mix notations as in (ai, bj, ck). (Unless, of course, your teacher does it that way- don't rock the boat!)

 

[inv]

I got it.Thx