1. perpendicular

a)Find the eqution of the plane that contains points (2,-1,4) , (3,2,-1)
and perpendicular to the plane x+y+2z=3

b)Find the equation of the plane containing the point (3,-1,-5) and perpendicular to the planes 3x-2y+2z = -7 and 5x-4y+3z=-1

2. plane

The vector normal to the plane $x+y+2z=3$is (1,1,2). You know it from the coefficients in front of x,y and z.
You know that this vector will be part of the perpendicular plane since it is the normal vector. Then you want the vector linking points (2,-1,4) , (3,2,-1) to be in the perpendicular plane as well.
Find this vector and compute the cross product with the normal vector (1,1,2). This will give you the normal vector of the perpendicular plane (say (a,b,c)).
The equation of the plane is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ where $x_0 y_0 z_0$ are the point of either of the points (2,-1,4) , (3,2,-1).

3. b)Find the equation of the plane containing the point (3,-1,-5) and perpendicular to the planes 3x-2y+2z = -7 and 5x-4y+3z=-1
$n_{1}=[3,-2,2]$

$n_{2}=[5,-4,3]$

Now, take the Cross product of the above normals and get [2,1,-2]

If we take $2x+y-2z$ and plug in the given point, we get 15

The equation of the plane is $2x+y-2z=15$

4. Originally Posted by Faz
a)Find the eqution of the plane that contains points (2,-1,4) , (3,2,-1)
and perpendicular to the plane x+y+2z=3

b)Find the equation of the plane containing the point (3,-1,-5) and perpendicular to the planes 3x-2y+2z = -7 and 5x-4y+3z=-1

plse assist...
For a plane, you need a point and a normal

a) use any of the two points given to get the point for a plane. u can use the cross product of the two line segments (which u find by using the points A=(2,-1,4) B=(3,2,-1) and C=(1,1,2) to get the normal. From there, u can find the equation.

b) its pretty much the same thing here...but this time to find the normal, u ll have to find the line segments by using the two points from the equations of the planes given and then use the cross product...

I hope it works...

5. for a)

you know that the new plane contains the normal vector to the other plane and the vector between the two points given.

Take the cross product between these two vectors to get the normal to the new plane.

Use one of the provided points and the normal vector to the new plane to get your equation for the plane.

I got, unsimplified, 11(x-2)-7(y+1)-2(z-4)=0

6. Originally Posted by Faz
a)Find the eqution of the plane that contains points (2,-1,4) , (3,2,-1)
and perpendicular to the plane x+y+2z=3

b)Find the equation of the plane containing the point (3,-1,-5) and perpendicular to the planes 3x-2y+2z = -7 and 5x-4y+3z=-1
Re-Hi !

a)The plane that contains points A(2,-1,4) , B(3,2,-1) and perpendicular to the plane x+y+2z=3 :
- also contains vector AB(1,3,-5)
- contains a perpendicular vector of plane x+y+2z=3, say u(1,1,2)

AB and u being not collinear are 2 direction vectors of the plane
Their product gives a vector perpendicular to the plane
$u \wedge {AB}$ (-11,7,2)

Therefore the plane equation is -11x+7y+2z+k=0
A being on the plane equation is -11x+7y+2z+21=0

b)the plane containing the point (3,-1,-5) and perpendicular to the planes 3x-2y+2z = -7 and 5x-4y+3z=-1 contains a perpendicular vector of the first one u(3,-2,2) and a perpendicular vector of the second one v(5,-4,3)

u and v are not collinear
$u \wedge v$ (2,1,-2) is a perpendicular vector of the plane
Therefore its equation is 2x+y-2z+k=0
The plane containing the point (3,-1,-5) => 2x+y-2z-15=0