# Find Plane Perpendicular to Vector & Pass Through Point

• starsiege
In summary, the book does not have an answer to the problem of finding the equation of a plane perpendicular to vector (2i+3j+4k) and passing through the point (1,2,3). Starsiege suggests using the scalar product or dot product to solve the problem. First, draw a line through the origin, O, parallel to (2i+3j+4k). Next, find the point, F, on the line which is the foot of the perpendicular from (1,2,3). Finally, find the condition for OF to be perpendicular to FP.
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 :)

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:
Draw the perpendicular!

starsiege said:
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?

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!)

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.

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)

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.

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.

## 1. What is the definition of a plane perpendicular to a vector?

A plane perpendicular to a vector is a flat surface that intersects the vector at a right angle, or 90 degrees. This means that the vector is perpendicular to every point on the plane and the plane is parallel to the direction of the vector.

## 2. How do you find a plane perpendicular to a given vector?

To find a plane perpendicular to a vector, you will need to use the cross product between the given vector and another vector. The resulting vector will be perpendicular to both the given vector and the new vector, and this will determine the direction of the plane. You can then use the equation of a plane to find the specific coordinates of the plane.

## 3. What is the equation of a plane passing through a given point?

The equation of a plane passing through a given point is represented as Ax + By + Cz = D, where A, B, and C are the coefficients of the variables x, y, and z, respectively. The values of A, B, and C are determined by the direction of the plane, and the value of D is determined by the coordinates of the given point.

## 4. Can a plane be perpendicular to more than one vector?

No, a plane can only be perpendicular to one vector. This is because a plane is defined by its normal vector, which is perpendicular to the plane. Therefore, if a plane is perpendicular to more than one vector, those vectors must be parallel to each other.

## 5. How can you verify that a plane is perpendicular to a vector?

To verify that a plane is perpendicular to a vector, you can use the dot product between the normal vector of the plane and the given vector. If the dot product is equal to zero, then the plane and the vector are perpendicular. Additionally, you can graph the plane and the vector to visually see if they intersect at a right angle.

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