line of intersection of planes

**ashes** · April 16th 2008, 01:02 AM

Hi..how do I find the unit vector parallel to the line of intersection of planes, 2x+2y-z=6 and x+2y+z=2. So far I got $2x+2y-z-6+\lambda(x+2y+z-2)=0$ ..
THen i need to find the parametric form of the equation of a line passing thru the point A[2,2,1] that is parallel to the line of the intersection of the 2 planes above.

THanks

**earboth** · April 16th 2008, 01:18 AM

Originally Posted by ashes

Hi..how do I find the unit vector parallel to the line of intersection of planes, 2x+2y-z=6 and x+2y+z=2. So far I got $2x+2y-z-6+\lambda(x+2y+z-2)=0$ ..
THen i need to find the parametric form of the equation of a line passing thru the point A[2,2,1] that is parallel to the line of the intersection of the 2 planes above.

THanks

The normal vectors of the planes are perpendicular to the direction of the line of intersection. Thus:

$\vec d = (2, 2, -1) \times (1, 2, 1) = (4, -3, 2)$

Then the equation of a line passing through A in the direction of $\vec d$ is:

$\vec r = (2, 2, 1) + s \cdot (4, -3, 2)$ ...... which will yield:

$l:\left\{ \begin{array}{lcr}x&=& 2+4s \\ y&=&2 -3s \\z&=& 1 + 2s\end{array} \right.$

**ashes** · April 16th 2008, 02:15 AM

ic..so thats a line of intersection of 2 planes.so the unit vector which is parallel to that line is $\frac{4}{\sqrt{29}},\frac{-3}{\sqrt{29}},\frac{2}{\sqrt{29}} ?$ so to deduce, any line which is parallel to another line will have the same direction vector and a point which lies on it.. so the equation of the line of intersection is [a,b,c]+s[4,-3,2]..since we are not given any points which is on that line..pls correct me if im wrong..

**earboth** · April 16th 2008, 03:39 AM

Originally Posted by ashes

ic..so thats a line of intersection of 2 planes.so the unit vector which is parallel to that line is $\frac{4}{\sqrt{29}},\frac{-3}{\sqrt{29}},\frac{2}{\sqrt{29}} ?$ ...... Right!

so to deduce, any line which is parallel to another line will have the same direction vector and a point which lies on it........ Right!

so the equation of the line of intersection is [a,b,c]+s[4,-3,2]..since we are not given any points which is on that line........ Right!

pls correct me if im wrong.. Not necessary! You got it.

..

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