Find Plane Perpendicular to Vector & Pass Through Point

[starsiege]

Hey guys; I am 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 :)

 

[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.

 

[tiny-tim]

Draw the perpendicular!

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?

 

[HallsofIvy]

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!)

 

[starsiege]

thank you everyone for your help and suggestions.!

i solved it...in fact I am 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.

 

[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)

 

[HallsofIvy]

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.

 

[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.