how to find the equation of a plane given two vectors

Lesson Explainer: Equation of a Plane: Vector, Scalar, and General Forms Mathematics

In this explainer, we will larn how to discover the vector, scalar (standard or component), and general (Cartesian or normal) forms of the equation of a aeroplane given the normal vector and a point on it.

Let's first consider the equation of a line in Cartesian form and rewrite it in vector form in ii dimensions, , as the situation will be similar for a plane in three dimensions, .

Recall that the general form of the equation of a directly line in 2 dimensions is

This tin also be written in the form , where is the gradient and is the -intercept, which we tin can determine past knowing two points on the line. If is a indicate that lies on the line, we can determine from the general form as thus, the equation of the line can be written every bit

The equation of the line can as well exist realized as a dot product of two vectors every bit

Now if we define the position vectors, then the equation of the line tin can be written in vector grade equally where is called a normal vector of the line and will lie completely on the line. A holding of the dot product states that two vectors are perpendicular to each other if their dot product is null. This equation of the line in vector grade shows that the normal vector and the vector are perpendicular to each other past this property.

A normal vector to a line or aeroplane is a vector that is perpendicular to the line or plane. In other words, the normal vector is perpendicular to any vector that is parallel to the line or plane, and nosotros take , by the property of the dot product.

Similar to the equation of a line in ii dimensions, the equation of a plane in three dimensions can be represented in terms of the normal vector on the airplane. We can stand for the equation of a plane as follows.

Definition: Full general Form of the Equation of a Plane

The general form of the equation of a plane in is where , , and are the components of the normal vector , which is perpendicular to the plane or any vector parallel to the aeroplane.

If is a point that lies on the aeroplane, then and nosotros can write the equation of the plane as

This tin can be rearranged to give the equation of the plane in scalar form.

Definition: Scalar Course of the Equation of a Plane

The scalar form of the equation of a plane in containing the bespeak is where , , and are the components of the normal vector , which is perpendicular to the plane or any vector parallel to the plane.

Now, let's consider an instance where nosotros make up one's mind the equation of the plane in this form from the normal vector and a given point that lies on the aeroplane.

Instance 1: Finding the Equation of a Airplane given a Point and Its Normal Vector

Give the equation of the plane with normal vector that contains the point .

Answer

In this instance, we want to determine the equation of the aeroplane by using one betoken on the plane and a given normal vector to the aeroplane.

Recall that the scalar form of the equation of a plane with a normal vector that contains the indicate is

Thus, substituting the values for the given normal vector and point , we have

Thus, the general form of the equation of the plane with normal vector that contains the betoken is

If nosotros are given a indicate that lies on the plane, , and two nonparallel vectors, and , that are parallel to the plane, and then we tin can determine the normal to the plane from these ii vectors. Since the vectors are both parallel to the aeroplane, the normal vector must be perpendicular to both and . Recall that the cross product of 2 vectors produces a vector that is perpendicular to both vectors. We can utilise this holding of the cantankerous product to compute a normal vector to the aeroplane, which leads to the normal vector

In the next case, we will determine the equation of the plane past get-go finding the normal vector of the plane from two vectors that are parallel to it.

Example 2: Finding the General Equation of a Aeroplane Passing through a Given Signal and Parallel to Two Given Vectors

Discover the general equation of the aeroplane that passes through the point and is parallel to the two vectors and .

Respond

In this example, we want to determine the equation of the plane that passes through a signal and is parallel to two given vectors.

Recall that the scalar form of the equation of a aeroplane with a normal vector that contains the indicate is

We need to decide a normal, , to the plane, which is a vector perpendicular to both and , since these are parallel to the aeroplane. Nosotros can notice the normal vector past taking the cantankerous product betwixt these vectors:

Using the normal vector and a point on the plane , we have

Thus, the general equation of the airplane that passes through the signal and is parallel to the ii vectors and is

If a plane contains three points , , and , and then nosotros tin can decide the equation of the plane. By substituting these points into the scalar course of the equation of the plane nosotros become like to how nosotros can make up one's mind the equation of a line with two given points. However, this is non the standard fashion to determine the equation of a aeroplane. Instead, we shall decide a normal vector by noting that the difference of the position vectors of any ii points on the plane is a vector parallel to the aeroplane; we shall revisit this when because the vector class of the equation of the plane.

If nosotros denote the position vectors of the three noncollinear points equally , , and , and so we tin can obtain two vectors parallel to the plane by subtracting pairs of these position vectors every bit

In fact, we can do this with any pairs and in any club; for example, another choice could be

In whatsoever case, the normal vector tin exist determined from the cross product of these two vectors:

Now, let's consider an example where nosotros use this class along with information well-nigh iii points that prevarication on the plane to determine the equation.

Example three: Finding the General Equation of a Aeroplane Passing through 3 Noncollinear Points

Write, in normal form, the equation of the plane , , and .

Answer

In this example, we desire to determine the equation of the airplane from three given points that lie on the airplane.

Think that the equation of a aeroplane with a normal vector that contains the bespeak is

Let's first make up one's mind a normal vector to the plane. We tin obtain two vectors in the plane past subtracting the position vectors of pairs of points on the plane:

We tin find the normal vector by taking the cantankerous product between these vectors:

Using the normal vector and any of the given points that prevarication on the plane, for example, , the equation of the plane becomes

Thus, dividing past x, we obtain the equation of the plane in general class as

The scalar equation of a airplane tin can besides be realized equally the dot production of 2 vectors as

Now, let be a bespeak on the airplane and exist whatever betoken on the plane, represented by the position vectors and respectively, that is, and , and let be a normal vector to the plane.

The equation of the airplane in the vector grade tin be written as

The vector is perpendicular to the aeroplane, which means it is perpendicular to the vector of the difference of position vectors of any ii points on the airplane. This makes sense considering, past structure, will e'er lie completely on the plane and the dot product of this vector with the normal vector is zero, which means they are perpendicular.

This equation of the plane can exist rearranged to requite the vector course of the equation of a plane.

Definition: Vector Form of the Equation of a Plane

The vector form of the equation of a plane in is where is the position vector of any bespeak that lies on the plane and is a normal vector that is perpendicular to the aeroplane or whatever vector parallel to the plane.

Now, let's await at two examples where we determine the equation of the planes in vector class from given normal vectors and points that lie on the plane.

Instance 4: Finding the Vector Form of the Equation of a Plane given Its Normal Vector Equation

Find the vector form of the equation of the airplane that has normal vector and contains the point .

Answer

In this example, we want to determine the equation of a plane, in vector form, past using a point that lies on the plane and a given normal vector.

Recall that the vector equation of the plane can be written every bit where is a normal vector to the airplane and is the position vector of a point that lies on the plane.

The equation of the plane with normal vector that contains the point with position vector is

Thus, the vector form of the equation of the plane is

At present, permit'southward consider an instance where we convert the equation of the plane from general form to vector class.

Example 5: Finding the Vector Grade of the Equation of a Aeroplane

The equation of a aeroplane has the general form . What is its vector form?

Answer

In this example, we want to decide the equation of the plane in vector class by using the given equation of the airplane in general form.

Recall that the general form of the equation of a plane in is where , , and are the components of the normal vector , which is perpendicular to the aeroplane or whatever vector parallel to the plane. The vector equation of the airplane can be written every bit

From the given equation of the plane, , we can identify the normal vector equally and . The vector equation of the aeroplane can exist written equally

The equation of a straight line in in vector form is where is a position vector of a bespeak on the line and is a vector parallel to the line.

If nosotros have a airplane containing 2 district intersecting directly lines with vector equations and then we tin determine a point that lies on the plane from either of these equations. For simplicity, we can substitute in the get-go equation, giving the position vector of a betoken that lies on the first line and, hence, the plane as .

In order to determine the normal vector to the plane, we note that the vectors and parallel to the lines and are both parallel to the plane. Hence, we demand to make up one's mind a vector that is perpendicular to both and in order to determine the normal. Every bit long as and are not parallel, we can obtain the normal vector to the plane by taking the cross production of the two vectors:

Past putting these together, the equation of the plane tin can be written as

In our terminal example, nosotros will determine the equation of a plane in vector form from the vector equations of 2 directly lines that lie on the plane.

Example vi: Finding the Vector Equation of a Plane Containing Two Lines given Their Vector Equations

Find the vector form of the equation of the plane containing the ii directly lines and .

Reply

In this example, we want to determine the equation of the plane that contains two straight lines whose equations are given in vector form.

The vector grade of the equation of the airplane tin be written as where is a normal vector to the plane and is the position vector of a signal that lies on the plane.

For simplicity, allow's begin past writing the vector equations of the direct lines as

We note that, from the vector equations of the direct lines, the vector is parallel to the first line and is parallel to the 2d, which ways both are parallel to the plane. Thus, in order to make up one's mind a normal vector, , to the plane, we need to find a vector that is perpendicular to both and . We can exercise this past taking the cross production:

We tin can determine the position vector for a point on the plane from either equation of the line, since both lines are contained on the plane. For simplicity, we can substitute in the first equation to determine the position vector of the point as

Substituting the normal vector and the position vector of a point on the aeroplane , we have

Thus, the equation of the plane containing the two direct lines and in vector form is

Key Points

• The full general grade of the equation of a aeroplane in is
• If is a indicate that lies on the plane, and so , and we tin write the bespeak–normal, or scalar, form of the equation of the airplane as where , , and are the components of the normal vector , which is perpendicular to the airplane or any vector parallel to the airplane.
• If we are given a point on the aeroplane and two nonzero and nonparallel vectors and , which are parallel to the plane, we can determine the normal vector from the cross product:
• The equation of a plane in vector form tin can be written as with and as the position vector of a signal that lies on the plane.

Source: https://www.nagwa.com/en/explainers/373101390857/

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