The equation of a plane perpendicular to a vector and passing through a given point?

1. Mar 2, 2008

starsiege

Hey guys; im studying calculus and i came across a problem for which the book does not have an answer...

how do i find the equation of a plane perpendicular to vector (2i+3j+4k) and passing though
the point (1,2,3)

^^the numbers are some i just made up and no this is not a homework q....in fact i would be glad if any of you can kindly explain to me how this problem can be solved and/or if they can give me the link to some webpage where they have stuff about this

thanks :)

2. Mar 2, 2008

Dmak

The scalar product or dot product, allows you to tell if two vectors are orthogonal or perpendicular among other things. So if you can visualize that the vector $$(2, 3, 4)^{T}$$ must be orthogonal to every point in the plane then the plane through the origin with normal vector $$(2, 3, 4)^{T}$$ would be given by the equation $$2x + 3y +4z = 0$$. This is evident because $$(2, 3, 4) \bullet ( x, y, z )^{T} = 2x + 3y +4z = 0$$ implies $$(2, 3, 4)^{T}$$ is orthogonal to every vector in the plane. Now I leave it to you to figure out how to shift the plane so that it goes through the specified point.

Last edited: Mar 2, 2008
3. Mar 2, 2008

tiny-tim

Draw the perpendicular!

Hi starsiege! Welcome to PF!

Hint: All lines in that plane will be perpendicular to that vector!

So draw the line through the origin, O, parallel to (2i+3j+4k).

You want to find the point, F, on that line which is the foot of the perpendicular from (1,2,3), which we'll call P.

Call F (2a,3a,4a). Then what is the condition for OF to be perpendicular to FP?

4. Mar 2, 2008

HallsofIvy

Staff Emeritus
If you are expected to be able to do a problem like this, you should already know two things:

1) A line in 3 dimensions can be written in parametric equations, x= At+ x0, y= Bt+ y0, z= Ct+ z0, where (x0, y0, z0) is point on the line and $A\vec{i}+ B\vec{j}+ C\vec{k}$ is parallel to the line.

2) A plane can be written as a single equation, A(x- x0)+ B(y- y0)+ C0(z- z0)= 0 where (x0, y0, z0) is a point on the plane and $A\vec{i}+ B\vec{j}+ C\vec{k}$ is perpendicular to the plane.

You are given $A\vec{i}+ B\vec{j}+ C\vec{k}$ and (x0, y0, z0).

(My mistake, you don't really need to know (1) to do this problem!)

5. Mar 2, 2008

starsiege

thank you everyone for your help and suggestions.!

i solved it....in fact im ashamed to see how easy it was...
but even then i might not have found the answer if not for ur help! thanks again

Ps: i really should stop doing math till 1 am :zzz: it took me tens of mins to even get my mind around problems at that time while it took less than a min in the morning.

6. Mar 2, 2008

starsiege

hehe ran into another prob as i was going through

how do i go about a problem that gives me 3 points of a triangle A,B,C [ A (x1,y1), B(x2,y2),C(x3,y3) ] and asks me to find the interior angles of it?

how should i start on this problem?(this is in the section that we use dot product)

7. Mar 2, 2008

HallsofIvy

Staff Emeritus
Have you learned that, in additon to the "add the products of the components" formula, $\vec{u}\cdot\vec{v}= ||\vec{u}||||\vec{v}|| cos(\theta)$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$? That should do it easily.

8. Mar 2, 2008

starsiege

hey , thanks :) i figured it out soon after i posted the q but was not able to delete my q right away cos i was away from the comp. but thanks again.