What are p and q in relation to the given plane and vector?

[joshwoods]

Homework Statement

c = 10i+2j-3k = p + q

Calculate p and q where the vector p is parallel and q is perpendicular to the following plane:

2x+3y-z=4

Homework Equations

a[itex]\bullet[/itex]b= axbx+ayby+azbz

The Attempt at a Solution

I know that because p is perpendicular to q the dot product between these two vectors should be zero. Also a vector parallel to the plane above is one which satisfies to equation of the plane and any vector perpendicular to the plane is a scalar multiple of the vector:

2x+3y-z

 

[HallsofIvy]

Good- though i would not write the vector as "2x+3y-z". You have already used x, y, and z as variables in the equation of the plane so it would better to use, say, i, j, and k as the unit vectors in the x, y, and z directions. That is, 2i+ 3j- k. Since 10i+2j-3k = p + q= p+ 2ai+ 3aj- ak, you can write p= (10- 2a)i+ (2- 3a)j+ (-1+ a)k. And, of course, because they are perpendicular, the dot product of p and q is 0: (10- 2a)(10)+ (2- 3a)(2)+ (-1+ a)(-1)= 0.

Solve that equation for a.

 

[joshwoods]

Hello,

Thanks for your help, I am still a little confused with the solution, when I solve for p I get:

p = (10-2a)i+(2-3a)j+(-3+a)k

but I am also confused about the dot product, should it not be:

q = 2ai+3aj-ak

pdotq = (10-2a)i(2a)i + (2-3a)j(3a)j + (-3+a)k(-a)k

This is what I get when I solve is out but my final answer still does not make sense. I must be making a small error at some point. It would be really great if you could help me out.

 

[ehild]

Homework Statement

c = 10i+2j-3k = p + q

Calculate p and q where the vector p is parallel and q is perpendicular to the following plane:

2x+3y-z=4

Homework Equations

a[itex]\bullet[/itex]b= axbx+ayby+azbz

The Attempt at a Solution

I know that because p is perpendicular to q the dot product between these two vectors should be zero. Also a vector parallel to the plane above is one which satisfies to equation of the plane and any vector perpendicular to the plane is a scalar multiple of the vector:

2x+3y-z

A vector lying in the plane does not satisfy the equation of the plane. The equation of the plane means that the difference between two vectors of the plane r and ro is perpendicular to the normal of the plane: (r-ro)n=0 which leads to the equation ax+bx+cz=C, where a,b,c, are the components of the normal and C is the dot product of ro with the normal vector.

The normal vector of the plane 2x+3y-z=4 is n=(2,3,-1), a vector, perpendicular to the plane. The vector c then is the sum of a vector, parallel to n and an other, perpendicular to n. You certainly know how to get the projection of a vector to a certain direction?

ehild