Finding the Intersection of Two Planes using Vector Equations

[PeterSK]

Homework Statement

Two planes r_1 and r_2 have the equations:

r_1 = ( 1 - \lambda ) \underline{i} + ( 2 \lambda + \mu ) \underline{j} + ( \mu - 1 ) \underline{k}

r_2 = ( s - t ) \underline{i} + ( 2s - 3 ) \underline{j} + ( t ) \underline{k}

If a point lies in both r_1 and r_2 then \mu =4 \lambda + 3 (shown in a previous question)

Hence find a vector equation of the line of intersection of the two planes.

Homework Equations

None known

The Attempt at a Solution

I know what I have to do but I have no idea how to do it:

• Find the normals of the planes
• Use the cross (vector) product on them to get the direction of the intersection vector
• find a point on the vector (I assume using the \mu = 4 \lambda + 3 stuff)
• substitute the two parts into the formula for a vector equation to get the answer

However, I have no idea how to find the normals of those planes and I can also see finding the point to be awkward too with all those mu's, lambda's, s's and t's.
I'm just completely stumped!

 

[andytoh]

Put the planes in the form Ax +By +C=0, eliminating the parameters in so doing, and the normals will be (A,B,C)

 

[PeterSK]

The only way I know of making that form is by doing the dot product of the plane and its normal which doesn't help as the normal is what I'm trying to find

 

[HallsofIvy]

If you have determined that \mu= 4\lambda + 3, that's all you need!

Replace \mu by 4\lambda + 3 it the equation for the first plane:
\vec{r_1}= (1- \lambda)\vec{i})+ (2\lambda + (4\lambda +3))\vec{j}+ ((4\lambda+ 3)- 1)\vec{k}
\vec{r_1}= (1-\lambda)\vec{i}+ (6\lambda+ 3)\vec{j}+ k+ (4\lambda+ 2)\vec{k}
That is the vector equation of the line satisfying \mu= 4\lambda+ 3j- which you say is true for any point on the line of intersection.

 

[PeterSK]

Thanks a lot, that helped loads!

 

[PeterSK]

Sorry, but is that just the direction vector of the line of intersection?
If so, then do I need to make it into the form r = \underline{a} + \lambda \underline{c} ?

 

[HallsofIvy]

No, that is not the direction vector, it is the position vector.

 

[PeterSK]

Great, thanks a lot.