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

by starsiege
Tags: equation, passing, perpendicular, plane, point, vector
 P: 6 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 :)
 P: 15 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.
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 Quote by starsiege how do i find the equation of a plane perpendicular to vector (2i+3j+4k) and passing though the point (1,2,3)
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?

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

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.

 how do i find the equation of a plane perpendicular to vector (2i+3j+4k) and passing though the point (1,2,3)
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!)
 P: 6 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 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.
 P: 6 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)
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,568 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.
 P: 6 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.

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