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how to find the equation of a plane given two vectors

Written By Friday, May 20, 2022 Add Comment Edit

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, ℝ , as the situation will be similar for a plane in three dimensions, ℝ .

Recall that the general form of the equation of a directly line in 2 dimensions is 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 = 0 .

This tin also be written in the form 𝑦 = 𝑚 𝑥 + 𝑑 , where 𝑚 is the gradient and 𝑑 is the 𝑦 -intercept, which we tin can determine past knowing two points on the line. If ( 𝑥 , 𝑦 ) is a indicate that lies on the line, we can determine 𝑐 from the general form as 𝑐 = − ( 𝑎 𝑥 + 𝑏 𝑦 ) ; thus, the equation of the line can be written every bit 𝑎 𝑥 + 𝑏 𝑦 − ( 𝑎 𝑥 + 𝑏 𝑦 ) = 0 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) = 0 .

The equation of the line can as well exist realized as a dot product of two vectors every bit ( 𝑎 , 𝑏 ) ⋅ ( 𝑥 − 𝑥 , 𝑦 − 𝑦 ) = 0 ( 𝑎 , 𝑏 ) ⋅ ( ( 𝑥 , 𝑦 ) − ( 𝑥 , 𝑦 ) ) = 0 .

Now if we define the position vectors, ⃑ 𝑟 = ( 𝑥 , 𝑦 ) , ⃑ 𝑟 = ( 𝑥 , 𝑦 ) , then the equation of the line tin can be written in vector grade equally ⃑ 𝑛 ⋅ ⃑ 𝑟 − ⃑ 𝑟 = 0 ⃑ 𝑛 ⋅ ⃑ 𝑟 = ⃑ 𝑛 ⋅ ⃑ 𝑟 , where ⃑ 𝑛 = ( 𝑎 , 𝑏 ) is called a normal vector of the line and ⃑ 𝑟 − ⃑ 𝑟 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 ⃑ 𝑛 and the vector ⃑ 𝑟 − ⃑ 𝑟 are perpendicular to each other past this property.

A normal vector ⃑ 𝑛 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 ⃑ 𝑣 that is parallel to the line or plane, and nosotros take ⃑ 𝑛 ⋅ ⃑ 𝑣 = 0 , 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 ℝ is 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 , where 𝑎 , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) , which is perpendicular to the plane or any vector parallel to the aeroplane.

If ( 𝑥 , 𝑦 , 𝑧 ) is a point that lies on the aeroplane, then 𝑑 = − ( 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 ) and nosotros can write the equation of the plane as 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 − ( 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 ) = 0 .

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 ℝ containing the bespeak ( 𝑥 , 𝑦 , 𝑧 ) is 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , where 𝑎 , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) , 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 ( 1 0 , 8 , iii ) that contains the point ( 1 0 , 5 , 5 ) .

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 ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) that contains the indicate ( 𝑥 , 𝑦 , 𝑧 ) is 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 .

Thus, substituting the values for the given normal vector ( ane 0 , 8 , 3 ) and point ( 1 0 , 5 , 5 ) , we have one 0 ( 𝑥 − 1 0 ) + eight ( 𝑦 − five ) + 3 ( 𝑧 − 5 ) = 0 1 0 𝑥 − 1 0 0 + eight 𝑦 − 4 0 + 3 𝑧 − 1 five = 0 1 0 𝑥 + eight 𝑦 + iii 𝑧 − one v v = 0 .

Thus, the general form of the equation of the plane with normal vector ( 1 0 , viii , 3 ) that contains the betoken ( 1 0 , 5 , five ) is 1 0 𝑥 + 8 𝑦 + 3 𝑧 − 1 v 5 = 0 .

If nosotros are given a indicate that lies on the plane, ( 𝑥 , 𝑦 , 𝑧 ) , and two nonparallel vectors, ⃑ 𝑣 and ⃑ 𝑣 , 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 ⃑ 𝑣 and ⃑ 𝑣 . 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 ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 .

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 ( v , 1 , − i ) and is parallel to the two vectors ( nine , seven , − 8 ) and ( − two , 2 , − one ) .

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 ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) that contains the indicate ( 𝑥 , 𝑦 , 𝑧 ) is 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 .

We need to decide a normal, ⃑ 𝑛 , to the plane, which is a vector perpendicular to both ( 9 , 7 , − 8 ) and ( − 2 , 2 , − 1 ) , since these are parallel to the aeroplane. Nosotros can notice the normal vector past taking the cantankerous product betwixt these vectors: ⃑ 𝑛 = ( 9 , seven , − 8 ) × ( − 2 , 2 , − i ) = | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 9 vii − 8 − 2 ii − 1 | | | | = | | seven − 8 two − ane | | ⃑ 𝑖 − | | nine − viii − 2 − 1 | | ⃑ 𝑗 + | | 9 7 − two 2 | | ⃑ 𝑘 = ( 7 × ( − ane ) − ( − eight ) × two ) ⃑ 𝑖 − ( 9 × ( − ane ) − ( − 8 ) × ( − 2 ) ) ⃑ 𝑗 + ( 9 × ii − 7 × ( − 2 ) ) ⃑ 𝑘 = ix ⃑ 𝑖 + 2 v ⃑ 𝑗 + 3 2 ⃑ 𝑘 = ( ix , 2 5 , 3 2 ) .

Using the normal vector ( ix , 2 5 , three 2 ) and a point on the plane ( 5 , 1 , − 1 ) , we have 9 ( 𝑥 − 5 ) + ii 5 ( 𝑦 − 1 ) + 3 2 ( 𝑧 + 1 ) = 0 9 𝑥 − 4 5 + 2 5 𝑦 − 2 v + 3 ii 𝑧 + 3 2 = 0 ix 𝑥 + 2 5 𝑦 + 3 2 𝑧 − 3 8 = 0 .

Thus, the general equation of the airplane that passes through the signal ( 5 , 1 , − 1 ) and is parallel to the ii vectors ( 9 , 7 , − 8 ) and ( − 2 , two , − ane ) is ix 𝑥 + 2 5 𝑦 + iii two 𝑧 − 3 eight = 0 .

If a plane contains three points ( 𝑥 , 𝑦 , 𝑧 ) , ( 𝑥 , 𝑦 , 𝑧 ) , and ( 𝑥 , 𝑦 , 𝑧 ) , 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 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , 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 ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) , ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) , and ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) , and so we tin can obtain two vectors parallel to the plane by subtracting pairs of these position vectors every bit ⃑ 𝑣 = ⃑ 𝑟 − ⃑ 𝑟 = ( 𝑥 − 𝑥 , 𝑦 − 𝑦 , 𝑧 − 𝑧 ) , ⃑ 𝑣 = ⃑ 𝑟 − ⃑ 𝑟 = ( 𝑥 − 𝑥 , 𝑦 − 𝑦 , 𝑧 − 𝑧 ) .

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

In whatsoever case, the normal vector tin exist determined from the cross product of these two vectors: ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 .

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 ( 1 , 0 , 3 ) , ( 1 , 2 , − ane ) , and ( half dozen , one , 6 ) .

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 ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) that contains the bespeak ( 𝑥 , 𝑦 , 𝑧 ) is 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 .

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: ⃑ 𝑣 = ( 1 , 0 , 3 ) − ( 1 , two , − 1 ) = ( 0 , − 2 , 4 ) , ⃑ 𝑣 = ( 1 , 0 , iii ) − ( 6 , ane , 6 ) = ( − 5 , − one , − iii ) .

We tin find the normal vector by taking the cantankerous product between these vectors: ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 = ( 0 , − ii , iv ) × ( − five , − 1 , − 3 ) = | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 0 − ii iv − v − 1 − 3 | | | | = | | − two 4 − ane − three | | ⃑ 𝑖 − | | 0 4 − 5 − 3 | | ⃑ 𝑗 + | | 0 − 2 − 5 − 1 | | ⃑ 𝑘 = ( ( − 2 ) × ( − three ) − 4 × ( − i ) ) ⃑ 𝑖 − ( 0 × ( − iii ) − iv × ( − five ) ) ⃑ 𝑗 + ( 0 × ( − 1 ) − ( − 2 ) × ( − 5 ) ) ⃑ 𝑘 = 1 0 ⃑ 𝑖 − 2 0 ⃑ 𝑗 − 1 0 ⃑ 𝑘 = ( one 0 , − 2 0 , − 1 0 ) .

Using the normal vector ( one 0 , − 2 0 , − 1 0 ) and any of the given points that prevarication on the plane, for example, ( i , 0 , 3 ) , the equation of the plane becomes i 0 ( 𝑥 − 1 ) − two 0 ( 𝑦 − 0 ) − one 0 ( 𝑧 − three ) = 0 1 0 𝑥 − 1 0 − 2 0 𝑦 − i 0 𝑧 + three 0 = 0 1 0 𝑥 − 2 0 𝑦 − 1 0 𝑧 + ii 0 = 0 .

Thus, dividing past x, we obtain the equation of the plane in general class as 𝑥 − 2 𝑦 − 𝑧 + 2 = 0 .

The scalar equation of a airplane tin can besides be realized equally the dot production of 2 vectors as ( 𝑎 , 𝑏 , 𝑐 ) ⋅ ( 𝑥 − 𝑥 , 𝑦 − 𝑦 , 𝑧 − 𝑧 ) = 0 ( 𝑎 , 𝑏 , 𝑐 ) ⋅ ( ( 𝑥 , 𝑦 , 𝑧 ) − ( 𝑥 , 𝑦 , 𝑧 ) ) = 0 .

Now, let 𝑃 = ( 𝑥 , 𝑦 , 𝑧 ) be a bespeak on the airplane and 𝑃 = ( 𝑥 , 𝑦 , 𝑧 ) exist whatever betoken on the plane, represented by the position vectors ⃑ 𝑟 and ⃑ 𝑟 respectively, that is, ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) and ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) , and let ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) be a normal vector to the plane.

The equation of the airplane in the vector grade tin be written as ⃑ 𝑛 ⋅ ⃑ 𝑟 − ⃑ 𝑟 = 0 .

The vector ⃑ 𝑛 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, ⃑ 𝑟 − ⃑ 𝑟 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 ℝ is ⃑ 𝑛 ⋅ ⃑ 𝑟 = ⃑ 𝑛 ⋅ ⃑ 𝑟 , where ⃑ 𝑟 is the position vector of any bespeak that lies on the plane and ⃑ 𝑛 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 ⃑ 𝑛 = ⃑ 𝑖 + ⃑ 𝑗 + ⃑ 𝑘 and contains the point ( 2 , half-dozen , six ) .

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 ⃑ 𝑛 ⋅ ⃑ 𝑟 = ⃑ 𝑛 ⋅ ⃑ 𝑟 , where ⃑ 𝑛 is a normal vector to the airplane and ⃑ 𝑟 is the position vector of a point that lies on the plane.

The equation of the plane with normal vector ⃑ 𝑛 = ( 1 , one , ane ) that contains the point ( 2 , 6 , 6 ) with position vector ⃑ 𝑟 = ( 2 , 6 , six ) is ( 1 , 1 , ane ) ⋅ ⃑ 𝑟 = ( 1 , ane , 1 ) ⋅ ( 2 , 6 , 6 ) = 1 × 2 + 1 × vi + ane × 6 = ane four .

Thus, the vector form of the equation of the plane is ( 1 , 1 , 1 ) ⋅ ⃑ 𝑟 = one 4 .

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 5 𝑥 + half dozen 𝑦 + 9 𝑧 − ii viii = 0 . 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 ℝ is 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 , where 𝑎 , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) , which is perpendicular to the aeroplane or whatever vector parallel to the plane. The vector equation of the airplane can be written every bit ⃑ 𝑛 ⋅ ⃑ 𝑟 = − 𝑑 .

From the given equation of the plane, 5 𝑥 + half dozen 𝑦 + nine 𝑧 − 2 8 = 0 , we can identify the normal vector equally ⃑ 𝑛 = ( v , vi , 9 ) and 𝑑 = − ii eight . The vector equation of the aeroplane can exist written equally ( 5 , 6 , 9 ) ⋅ ⃑ 𝑟 = two 8 .

The equation of a straight line in ℝ in vector form is ⃑ 𝑟 = ⃑ 𝑟 + 𝑡 ⃑ 𝑣 , 𝑡 ∈ ℝ , l i n due east where ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) is a position vector of a bespeak 𝑃 = ( 𝑥 , 𝑦 , 𝑧 ) on the line and ⃑ 𝑣 = ( 𝑎 , 𝑏 , 𝑐 ) is a vector parallel to the line.

If nosotros have a airplane containing 2 district intersecting directly lines with vector equations ⃑ 𝑟 = ⃑ 𝑎 + 𝑡 ⃑ 𝑣 , ⃑ 𝑟 = ⃑ 𝑎 + 𝑡 ⃑ 𝑣 , and then we tin determine a point that lies on the plane from either of these equations. For simplicity, we can substitute 𝑡 = 0 in the get-go equation, giving the position vector of a betoken that lies on the first line and, hence, the plane as ⃑ 𝑎 .

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

Past putting these together, the equation of the plane tin can be written as ⃑ 𝑣 × ⃑ 𝑣 ⋅ ⃑ 𝑟 = ⃑ 𝑣 × ⃑ 𝑣 ⋅ ⃑ 𝑎 .

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 ⃑ 𝑟 = ⃑ 𝑖 − ⃑ 𝑗 − three ⃑ 𝑘 + 𝑡 3 ⃑ 𝑖 + iii ⃑ 𝑗 + four ⃑ 𝑘 and ⃑ 𝑟 = − ⃑ 𝑖 − two ⃑ 𝑗 − 3 ⃑ 𝑘 + 𝑡 − ⃑ 𝑖 − 2 ⃑ 𝑗 − 4 ⃑ 𝑘 .

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 ⃑ 𝑛 ⋅ ⃑ 𝑟 = ⃑ 𝑛 ⋅ ⃑ 𝑟 , where ⃑ 𝑛 is a normal vector to the plane and ⃑ 𝑟 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 ⃑ 𝑟 = ( one , − 1 , − three ) + 𝑡 ( three , three , 4 ) , ⃑ 𝑟 = ( − one , − 2 , − 3 ) + 𝑡 ( − 1 , − 2 , − 4 ) .

We note that, from the vector equations of the direct lines, the vector ( 3 , 3 , 4 ) is parallel to the first line and ( − 1 , − 2 , − 4 ) 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, ⃑ 𝑛 , to the plane, we need to find a vector that is perpendicular to both ( − one , − 2 , − 4 ) and ( 3 , 3 , 4 ) . We can exercise this past taking the cross production: ⃑ 𝑛 = ( − 1 , − two , − 4 ) × ( three , 3 , 4 ) = | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ 𝑘 − i − ii − 4 iii three 4 | | | | = | | − 2 − iv 3 four | | ⃑ 𝑖 − | | − 1 − 4 three 4 | | ⃑ 𝑗 + | | − 1 − 2 iii 3 | | ⃑ 𝑘 = ( − ii × four + iv × 3 ) ⃑ 𝑖 − ( − one × four + 4 × 3 ) ⃑ 𝑗 + ( − ane × 3 + 2 × 3 ) ⃑ 𝑘 = four ⃑ 𝑖 − 8 ⃑ 𝑗 + 3 ⃑ 𝑘 = ( iv , − 8 , three ) .

We tin can determine the position vector ⃑ 𝑟 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 𝑡 = 0 in the first equation to determine the position vector of the point as ⃑ 𝑟 = ( 1 , − 1 , − 3 ) .

Substituting the normal vector ( four , − 8 , three ) and the position vector of a point on the aeroplane ( 1 , − ane , − 3 ) , we have ( four , − eight , 3 ) ⋅ ⃑ 𝑟 = ( 4 , − 8 , 3 ) ⋅ ( i , − ane , − 3 ) = iv × ane + ( − viii ) × ( − 1 ) + 3 × ( − 3 ) = three .

Thus, the equation of the plane containing the two direct lines ⃑ 𝑟 and ⃑ 𝑟 in vector form is ( 4 , − 8 , 3 ) ⋅ ⃑ 𝑟 = 3 .

Key Points

  • The full general grade of the equation of a aeroplane in ℝ is 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 .
  • If ( 𝑥 , 𝑦 , 𝑧 ) is a indicate that lies on the plane, and so 𝑑 = − ( 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 ) , and we tin write the bespeak–normal, or scalar, form of the equation of the airplane as 𝑎 ( 𝑥 − 𝑥 ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , where 𝑎 , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( 𝑎 , 𝑏 , 𝑐 ) , which is perpendicular to the airplane or any vector parallel to the airplane.
  • If we are given a point on the aeroplane ( 𝑥 , 𝑦 , 𝑧 ) and two nonzero and nonparallel vectors ⃑ 𝑣 and ⃑ 𝑣 , which are parallel to the plane, we can determine the normal vector from the cross product: ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 .
  • The equation of a plane in vector form tin can be written as ⃑ 𝑛 ⋅ ⃑ 𝑟 = ⃑ 𝑛 ⋅ ⃑ 𝑟 , with ⃑ 𝑟 = ( 𝑥 , 𝑦 , 𝑧 ) and ⃑ 𝑟 as the position vector of a signal that lies on the plane.

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

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