In view of the coronavirus pandemic, we are making LIVE CLASSES and VIDEO CLASSES completely FREE to prevent interruption in studies
Maths > Three Dimensional Geometry > Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
Three Dimensional Geometry

Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point

What is the equation of a plane that passes through a given point and is perpendicular to a given vector ]? In the three-dimensional space, a vector can pass through multiple planes but there will be one and only one plane to which the line will be normal and which passes through the given point. This lesson explains how to derive the equation in vector form and cartesian form, with a solved example at the end for your understanding.

Suggested Videos

Play
Play
Play
Arrow
Arrow
ArrowArrow
Distance Formula and Its Use in 3D Geometry
Section formula in 3D
Collinearity of three points in 3D
Slider

 

Derivation of the Equation in Vector form

Let us consider a plane which is perpendicular to the given vector \( \vec{n} \) and passes through the given point A. Here, we shall denote the position vector of the point \( A (x_1, y_1, z_1) \) by \( \vec{a} \). Let \( \vec{r} \) be the position vector of a point P (x, y, z) that lies on the plane. Now, on careful analysis, we can understand that the vector \( \vec{AP} \) is perpendicular to the vector normal to the plane under consideration i.e. \( \vec{n} \)

Browse more Topics Under Three Dimensional Geometry

In other words,

\( \vec{AP} . \vec{n} = 0 \)  ­­­­­­­­­……(1)

Note that the points O (origin), A and P form a triangle. We shall apply the triangle law of addition of vectors to obtain:

$$ \vec{OA} + \vec{AP} = \vec{OP} $$

i.e. \( \vec{AP} = \vec{r} – \vec{a} \)

By replacing the above derivation in equation (1) we get

$$ ( \vec{r} – \vec{a} ) . \vec{n} = 0 $$  which is the required equation of a plane passing through a given point and perpendicular to a given vector – in Vector form.

Derivation of the Equation in Cartesian Form

Cartesian

It is easy to derive the Cartesian equation of a plane passing through a given point and perpendicular to a given vector from the Vector equation itself. Let the given point be \( A (x_1, y_1, z_1) \) and the vector which is normal to the plane be ax + by + cz. Let P (x, y, z) be another point on the plane. Then, we have

$$ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}  $$

Or, $$ \vec{a} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}  $$

$$ \vec{n} = ax + by + cz $$

By using the vector equation and replacing, we have

$$ ( ( x \hat{i} + y \hat{j} + z \hat{k} ) – (x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k} ) . ( ax + by + cz ) = 0 $$

Simplifying the above, we can write the equation as

$$ a ( x – x_1) + b (y – y_1) + c (z – z_1) = 0  $$ is the required equation.

To get a better idea of how to solve problems on this, you can refer to the solved example below.

Solved Example for You

Question 1: Find the equation of the plane that passes through the point (1, 2, 3) and containing the normal vector \( \vec{n} = 15 \hat{i} + 9 \hat/{j} – 12 \hat{k} \)

Answer : Here, the given point is (1, 2, 3) so \( x_1 = 1 , y_1 = 2 , z_1 = 3 \)
Also, a = 15, b = 9 , c = – 12
Thus, the required equation of the plane is
15 (x – 1) + 9 (y – 2) – 12(z – 3) = 0
i.e. 15x + 9y – 12z = – 3
i.e. 5x + 3y – 4z = -1 is the answer.

Question 2: The plane which passes through the point (-1, 0, -6  ) and perpendicular to the line whose d.r’s are ( 6, 20, -1) also passes through the point

  1. (1, 1, -26  )
  2. (0, 0, 0  )
  3. (2, 1, -32  )
  4. (-1, 0, -6  )

Answer : B. The d.r’s of the perpendicular line give the normal. So the plane will be of the form, 6x – 20y + z = d. So it passes through (-1, 0, -6  )d = 0. Hence the plane passes through the origin.

Question 3: What is meant by Cartesian plane?

Answer: A Cartesian plane is described by two perpendicular number lines: the x-axis, and the y-axis. Similarly, the x-axis is horizontal and the y-axis is vertical. By making use of these axes, one can describe any point in the plane by making use of an ordered pair of numbers.

Question 4: Why is it called Cartesian?

Answer: They are referred to as Cartesian because the idea was created by the mathematician and philosopher Rene Descartes. She was also referred to as Cartesius. Moreover, he is also famous for saying the popular quote “I think, therefore I am”.

Question 5: What are Cartesian coordinates used for?

Answer: In mathematics, we make use of the Cartesian coordinate system to distinctively determine each point in the plane through two numbers, which we usually refer to as the x-coordinate and the y-coordinate of the point.

Question 6: Who created the Cartesian coordinate system?

Answer: René Descartes invented the Cartesian coordinate system which we also call the Cartesian system. He was a French mathematician who created it in the 17th century.

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

Leave a Reply

avatar
  Subscribe  
Notify of

Get Question Papers of Last 10 Years

Which class are you in?
No thanks.