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how to find the normal vector of a plane

Written By Thursday, May 12, 2022 Add Comment Edit

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

In this explainer, we volition acquire how to find the vector, scalar (standard or component), and general (Cartesian or normal) forms of the equation of a plane given the normal vector and a point on it.

Permit'due south start consider the equation of a line in Cartesian course and rewrite it in vector form in two dimensions, ℝ , every bit the state of affairs will exist similar for a plane in three dimensions, ℝ .

Recall that the general class of the equation of a straight line in 2 dimensions is π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 = 0 .

This tin as well be written in the form 𝑦 = π‘š π‘₯ + 𝑑 , where π‘š is the gradient and 𝑑 is the 𝑦 -intercept, which we can make up one's mind by knowing two points on the line. If ( π‘₯ , 𝑦 ) is a point that lies on the line, we can make up one's mind 𝑐 from the general form as 𝑐 = − ( π‘Ž π‘₯ + 𝑏 𝑦 ) ; thus, the equation of the line can be written as π‘Ž π‘₯ + 𝑏 𝑦 − ( π‘Ž π‘₯ + 𝑏 𝑦 ) = 0 π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) = 0 .

The equation of the line can also exist realized equally a dot production of two vectors equally ( π‘Ž , 𝑏 ) ⋅ ( π‘₯ − π‘₯ , 𝑦 − 𝑦 ) = 0 ( π‘Ž , 𝑏 ) ⋅ ( ( π‘₯ , 𝑦 ) − ( π‘₯ , 𝑦 ) ) = 0 .

At present if we define the position vectors, ⃑ π‘Ÿ = ( π‘₯ , 𝑦 ) , ⃑ π‘Ÿ = ( π‘₯ , 𝑦 ) , and so the equation of the line tin can be written in vector form as ⃑ 𝑛 ⋅ ⃑ π‘Ÿ − ⃑ π‘Ÿ = 0 ⃑ 𝑛 ⋅ ⃑ π‘Ÿ = ⃑ 𝑛 ⋅ ⃑ π‘Ÿ , where ⃑ 𝑛 = ( π‘Ž , 𝑏 ) is called a normal vector of the line and ⃑ π‘Ÿ − ⃑ π‘Ÿ will lie completely on the line. A property of the dot product states that two vectors are perpendicular to each other if their dot product is zero. This equation of the line in vector form shows that the normal vector ⃑ 𝑛 and the vector ⃑ π‘Ÿ − ⃑ π‘Ÿ are perpendicular to each other by this property.

A normal vector ⃑ 𝑛 to a line or airplane 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 airplane, and nosotros accept ⃑ 𝑛 ⋅ ⃑ 𝑣 = 0 , by the property of the dot production.

Similar to the equation of a line in two dimensions, the equation of a plane in three dimensions can be represented in terms of the normal vector on the plane. We can correspond the equation of a plane as follows.

Definition: Full general Course of the Equation of a Airplane

The general form of the equation of a plane in ℝ is π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 , where π‘Ž , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) , which is perpendicular to the plane or any vector parallel to the plane.

If ( π‘₯ , 𝑦 , 𝑧 ) is a point that lies on the airplane, then 𝑑 = − ( π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 ) and we tin can write the equation of the airplane as π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 − ( π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 ) = 0 .

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

Definition: Scalar Form of the Equation of a Plane

The scalar form of the equation of a airplane in ℝ containing the bespeak ( π‘₯ , 𝑦 , 𝑧 ) is π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , where π‘Ž , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) , which is perpendicular to the plane or whatever vector parallel to the plane.

Now, let'southward consider an example where nosotros determine the equation of the plane in this grade from the normal vector and a given point that lies on the aeroplane.

Example 1: Finding the Equation of a Plane given a Point and Its Normal Vector

Requite the equation of the plane with normal vector ( 1 0 , viii , 3 ) that contains the point ( ane 0 , five , v ) .

Answer

In this example, we want to determine the equation of the airplane past using i signal 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 = ( π‘Ž , 𝑏 , 𝑐 ) that contains the indicate ( π‘₯ , 𝑦 , 𝑧 ) is π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 .

Thus, substituting the values for the given normal vector ( i 0 , viii , 3 ) and bespeak ( one 0 , 5 , five ) , we accept one 0 ( π‘₯ − 1 0 ) + 8 ( 𝑦 − 5 ) + 3 ( 𝑧 − 5 ) = 0 1 0 π‘₯ − 1 0 0 + 8 𝑦 − iv 0 + 3 𝑧 − 1 5 = 0 1 0 π‘₯ + eight 𝑦 + 3 𝑧 − i 5 5 = 0 .

Thus, the general form of the equation of the plane with normal vector ( 1 0 , eight , 3 ) that contains the point ( 1 0 , v , v ) is 1 0 π‘₯ + 8 𝑦 + 3 𝑧 − 1 five five = 0 .

If nosotros are given a point that lies on the plane, ( π‘₯ , 𝑦 , 𝑧 ) , and two nonparallel vectors, ⃑ 𝑣 and ⃑ 𝑣 , that are parallel to the plane, then we tin can determine the normal to the plane from these two vectors. Since the vectors are both parallel to the plane, the normal vector must be perpendicular to both ⃑ 𝑣 and ⃑ 𝑣 . Retrieve that the cross product of ii vectors produces a vector that is perpendicular to both vectors. Nosotros can use this property of the cross product to compute a normal vector to the plane, which leads to the normal vector ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 .

In the next example, we will make up one's mind the equation of the airplane by offset finding the normal vector of the plane from two vectors that are parallel to information technology.

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

Notice the full general equation of the plane that passes through the point ( 5 , ane , − one ) and is parallel to the two vectors ( 9 , 7 , − viii ) and ( − 2 , two , − 1 ) .

Answer

In this case, we want to make up one's mind the equation of the plane that passes through a point and is parallel to two given vectors.

Recall that the scalar form of the equation of a plane with a normal vector ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) that contains the bespeak ( π‘₯ , 𝑦 , 𝑧 ) is π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 .

We need to determine a normal, ⃑ 𝑛 , to the plane, which is a vector perpendicular to both ( nine , 7 , − 8 ) and ( − 2 , 2 , − 1 ) , since these are parallel to the plane. We can find the normal vector by taking the cross product betwixt these vectors: ⃑ 𝑛 = ( 9 , seven , − 8 ) × ( − 2 , 2 , − one ) = | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ π‘˜ 9 vii − 8 − two 2 − ane | | | | = | | 7 − 8 2 − 1 | | ⃑ 𝑖 − | | 9 − viii − 2 − 1 | | ⃑ 𝑗 + | | 9 seven − 2 2 | | ⃑ π‘˜ = ( 7 × ( − 1 ) − ( − eight ) × 2 ) ⃑ 𝑖 − ( nine × ( − 1 ) − ( − 8 ) × ( − two ) ) ⃑ 𝑗 + ( nine × 2 − seven × ( − 2 ) ) ⃑ π‘˜ = 9 ⃑ 𝑖 + 2 v ⃑ 𝑗 + iii 2 ⃑ π‘˜ = ( 9 , two 5 , three 2 ) .

Using the normal vector ( 9 , 2 five , three 2 ) and a betoken on the airplane ( five , one , − 1 ) , we have nine ( π‘₯ − 5 ) + 2 5 ( 𝑦 − 1 ) + 3 2 ( 𝑧 + 1 ) = 0 9 π‘₯ − four 5 + two v 𝑦 − 2 5 + 3 two 𝑧 + 3 2 = 0 9 π‘₯ + 2 5 𝑦 + iii ii 𝑧 − 3 viii = 0 .

Thus, the general equation of the plane that passes through the point ( 5 , i , − one ) and is parallel to the ii vectors ( 9 , seven , − 8 ) and ( − two , 2 , − one ) is ix π‘₯ + ii 5 𝑦 + 3 2 𝑧 − 3 8 = 0 .

If a plane contains three points ( π‘₯ , 𝑦 , 𝑧 ) , ( π‘₯ , 𝑦 , 𝑧 ) , and ( π‘₯ , 𝑦 , 𝑧 ) , then we can determine the equation of the plane. By substituting these points into the scalar form of the equation of the plane nosotros go π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , like to how we can determine the equation of a line with two given points. However, this is non the standard way to determine the equation of a plane. Instead, we shall determine a normal vector by noting that the deviation of the position vectors of any two points on the aeroplane is a vector parallel to the aeroplane; nosotros shall revisit this when considering the vector form of the equation of the plane.

If nosotros denote the position vectors of the three noncollinear points equally ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) , ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) , and ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) , and so we tin can obtain two vectors parallel to the plane by subtracting pairs of these position vectors as ⃑ 𝑣 = ⃑ π‘Ÿ − ⃑ π‘Ÿ = ( π‘₯ − π‘₯ , 𝑦 − 𝑦 , 𝑧 − 𝑧 ) , ⃑ 𝑣 = ⃑ π‘Ÿ − ⃑ π‘Ÿ = ( π‘₯ − π‘₯ , 𝑦 − 𝑦 , 𝑧 − 𝑧 ) .

In fact, we can practise this with any pairs and in any club; for example, some other choice could be ⃑ 𝑣 = ⃑ π‘Ÿ − ⃑ π‘Ÿ , ⃑ 𝑣 = ⃑ π‘Ÿ − ⃑ π‘Ÿ .

In any case, the normal vector tin be adamant from the cantankerous product of these two vectors: ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 .

Now, let'due south consider an instance where we employ this form along with information about three points that lie on the plane to decide the equation.

Example 3: Finding the Full general Equation of a Airplane Passing through Three Noncollinear Points

Write, in normal form, the equation of the plane ( 1 , 0 , 3 ) , ( 1 , 2 , − 1 ) , and ( 6 , 1 , 6 ) .

Answer

In this example, we want to determine the equation of the plane from three given points that prevarication on the plane.

Call up that the equation of a plane with a normal vector ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) that contains the betoken ( π‘₯ , 𝑦 , 𝑧 ) is π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 .

Allow's showtime determine a normal vector to the airplane. We can obtain two vectors in the plane by subtracting the position vectors of pairs of points on the plane: ⃑ 𝑣 = ( i , 0 , 3 ) − ( 1 , ii , − 1 ) = ( 0 , − 2 , iv ) , ⃑ 𝑣 = ( 1 , 0 , 3 ) − ( half-dozen , one , vi ) = ( − 5 , − 1 , − 3 ) .

Nosotros can find the normal vector past taking the cantankerous production between these vectors: ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 = ( 0 , − two , 4 ) × ( − v , − ane , − three ) = | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ π‘˜ 0 − two 4 − 5 − ane − iii | | | | = | | − 2 iv − 1 − iii | | ⃑ 𝑖 − | | 0 four − 5 − 3 | | ⃑ 𝑗 + | | 0 − 2 − five − one | | ⃑ π‘˜ = ( ( − two ) × ( − three ) − iv × ( − 1 ) ) ⃑ 𝑖 − ( 0 × ( − 3 ) − 4 × ( − 5 ) ) ⃑ 𝑗 + ( 0 × ( − i ) − ( − 2 ) × ( − 5 ) ) ⃑ π‘˜ = 1 0 ⃑ 𝑖 − 2 0 ⃑ 𝑗 − 1 0 ⃑ π‘˜ = ( one 0 , − 2 0 , − 1 0 ) .

Using the normal vector ( ane 0 , − two 0 , − 1 0 ) and any of the given points that prevarication on the plane, for instance, ( 1 , 0 , iii ) , the equation of the plane becomes ane 0 ( π‘₯ − 1 ) − ii 0 ( 𝑦 − 0 ) − 1 0 ( 𝑧 − 3 ) = 0 1 0 π‘₯ − ane 0 − 2 0 𝑦 − 1 0 𝑧 + 3 0 = 0 ane 0 π‘₯ − two 0 𝑦 − one 0 𝑧 + 2 0 = 0 .

Thus, dividing by 10, we obtain the equation of the plane in general form every bit π‘₯ − 2 𝑦 − 𝑧 + 2 = 0 .

The scalar equation of a plane can also be realized as the dot product of two vectors as ( π‘Ž , 𝑏 , 𝑐 ) ⋅ ( π‘₯ − π‘₯ , 𝑦 − 𝑦 , 𝑧 − 𝑧 ) = 0 ( π‘Ž , 𝑏 , 𝑐 ) ⋅ ( ( π‘₯ , 𝑦 , 𝑧 ) − ( π‘₯ , 𝑦 , 𝑧 ) ) = 0 .

Now, let 𝑃 = ( π‘₯ , 𝑦 , 𝑧 ) be a signal on the plane and 𝑃 = ( π‘₯ , 𝑦 , 𝑧 ) exist any point on the plane, represented by the position vectors ⃑ π‘Ÿ and ⃑ π‘Ÿ respectively, that is, ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) and ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) , and let ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) be a normal vector to the plane.

The equation of the plane in the vector form can exist written every bit ⃑ 𝑛 ⋅ ⃑ π‘Ÿ − ⃑ π‘Ÿ = 0 .

The vector ⃑ 𝑛 is perpendicular to the plane, which means it is perpendicular to the vector of the difference of position vectors of whatsoever 2 points on the plane. This makes sense considering, by structure, ⃑ π‘Ÿ − ⃑ π‘Ÿ will always prevarication completely on the aeroplane and the dot product of this vector with the normal vector is nil, which means they are perpendicular.

This equation of the plane can be rearranged to give the vector class of the equation of a aeroplane.

Definition: Vector Form of the Equation of a Airplane

The vector class of the equation of a aeroplane in ℝ is ⃑ 𝑛 ⋅ ⃑ π‘Ÿ = ⃑ 𝑛 ⋅ ⃑ π‘Ÿ , where ⃑ π‘Ÿ is the position vector of whatever signal that lies on the plane and ⃑ 𝑛 is a normal vector that is perpendicular to the plane or any vector parallel to the plane.

Now, allow'due south look at two examples where nosotros make up one's mind the equation of the planes in vector form from given normal vectors and points that prevarication on the plane.

Example 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 aeroplane that has normal vector ⃑ 𝑛 = ⃑ 𝑖 + ⃑ 𝑗 + ⃑ π‘˜ and contains the indicate ( 2 , 6 , 6 ) .

Answer

In this example, nosotros want to determine the equation of a plane, in vector course, by using a signal that lies on the airplane and a given normal vector.

Call up that the vector equation of the airplane tin be written equally ⃑ 𝑛 ⋅ ⃑ π‘Ÿ = ⃑ 𝑛 ⋅ ⃑ π‘Ÿ , where ⃑ 𝑛 is a normal vector to the plane and ⃑ π‘Ÿ is the position vector of a point that lies on the plane.

The equation of the plane with normal vector ⃑ 𝑛 = ( 1 , ane , one ) that contains the bespeak ( 2 , 6 , 6 ) with position vector ⃑ π‘Ÿ = ( 2 , 6 , 6 ) is ( ane , 1 , one ) ⋅ ⃑ π‘Ÿ = ( 1 , 1 , 1 ) ⋅ ( 2 , half dozen , 6 ) = one × 2 + 1 × 6 + 1 × 6 = one 4 .

Thus, the vector form of the equation of the plane is ( 1 , one , i ) ⋅ ⃑ π‘Ÿ = 1 4 .

Now, let's consider an case where we catechumen the equation of the plane from general form to vector grade.

Instance five: Finding the Vector Form of the Equation of a Plane

The equation of a plane has the general form 5 π‘₯ + vi 𝑦 + ix 𝑧 − ii 8 = 0 . What is its vector form?

Reply

In this case, we desire to decide the equation of the plane in vector course by using the given equation of the plane in general class.

Recollect that the general form of the equation of a plane in ℝ is π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 , where π‘Ž , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) , which is perpendicular to the plane or any vector parallel to the plane. The vector equation of the airplane can exist written equally ⃑ 𝑛 ⋅ ⃑ π‘Ÿ = − 𝑑 .

From the given equation of the plane, v π‘₯ + 6 𝑦 + 9 𝑧 − 2 viii = 0 , nosotros tin can identify the normal vector as ⃑ 𝑛 = ( v , vi , nine ) and 𝑑 = − 2 viii . The vector equation of the plane can be written equally ( 5 , 6 , ix ) ⋅ ⃑ π‘Ÿ = two viii .

The equation of a straight line in ℝ in vector form is ⃑ π‘Ÿ = ⃑ π‘Ÿ + 𝑑 ⃑ 𝑣 , 𝑑 ∈ ℝ , fifty i n e where ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) is a position vector of a indicate 𝑃 = ( π‘₯ , 𝑦 , 𝑧 ) on the line and ⃑ 𝑣 = ( π‘Ž , 𝑏 , 𝑐 ) is a vector parallel to the line.

If we have a plane containing two district intersecting directly lines with vector equations ⃑ π‘Ÿ = ⃑ π‘Ž + 𝑑 ⃑ 𝑣 , ⃑ π‘Ÿ = ⃑ π‘Ž + 𝑑 ⃑ 𝑣 , and so we can determine a point that lies on the plane from either of these equations. For simplicity, we tin can substitute 𝑑 = 0 in the first equation, giving the position vector of a point that lies on the first line and, hence, the airplane every bit ⃑ π‘Ž .

In order to decide the normal vector to the airplane, nosotros notation that the vectors ⃑ 𝑣 and ⃑ 𝑣 parallel to the lines ⃑ π‘Ÿ and ⃑ π‘Ÿ are both parallel to the plane. Hence, nosotros need to decide a vector that is perpendicular to both ⃑ 𝑣 and ⃑ 𝑣 in order to determine the normal. As long as ⃑ 𝑣 and ⃑ 𝑣 are not parallel, we tin can obtain the normal vector to the plane past taking the cross product of the two vectors: ⃑ 𝑛 = ⃑ 𝑣 × ⃑ 𝑣 .

By putting these together, the equation of the plane tin can be written equally ⃑ 𝑣 × ⃑ 𝑣 ⋅ ⃑ π‘Ÿ = ⃑ 𝑣 × ⃑ 𝑣 ⋅ ⃑ π‘Ž .

In our last example, nosotros volition determine the equation of a plane in vector class from the vector equations of two direct lines that prevarication on the plane.

Example 6: Finding the Vector Equation of a Plane Containing 2 Lines given Their Vector Equations

Find the vector grade of the equation of the airplane containing the two directly lines ⃑ π‘Ÿ = ⃑ 𝑖 − ⃑ 𝑗 − 3 ⃑ π‘˜ + 𝑑 iii ⃑ 𝑖 + 3 ⃑ 𝑗 + 4 ⃑ π‘˜ and ⃑ π‘Ÿ = − ⃑ 𝑖 − two ⃑ 𝑗 − three ⃑ π‘˜ + 𝑑 − ⃑ 𝑖 − ii ⃑ 𝑗 − 4 ⃑ π‘˜ .

Answer

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

The vector form of the equation of the plane can be written every bit ⃑ 𝑛 ⋅ ⃑ π‘Ÿ = ⃑ 𝑛 ⋅ ⃑ π‘Ÿ , where ⃑ 𝑛 is a normal vector to the plane and ⃑ π‘Ÿ is the position vector of a indicate that lies on the plane.

For simplicity, let's begin past writing the vector equations of the straight lines as ⃑ π‘Ÿ = ( i , − 1 , − three ) + 𝑑 ( 3 , 3 , four ) , ⃑ π‘Ÿ = ( − i , − two , − 3 ) + 𝑑 ( − ane , − two , − 4 ) .

Nosotros note that, from the vector equations of the straight lines, the vector ( 3 , iii , 4 ) is parallel to the starting time line and ( − 1 , − 2 , − iv ) is parallel to the second, which means both are parallel to the plane. Thus, in gild to determine a normal vector, ⃑ 𝑛 , to the plane, we demand to observe a vector that is perpendicular to both ( − 1 , − 2 , − iv ) and ( 3 , iii , four ) . We tin do this by taking the cross production: ⃑ 𝑛 = ( − ane , − 2 , − 4 ) × ( 3 , 3 , 4 ) = | | | | ⃑ 𝑖 ⃑ 𝑗 ⃑ π‘˜ − 1 − 2 − 4 3 3 4 | | | | = | | − ii − 4 3 4 | | ⃑ 𝑖 − | | − ane − 4 3 4 | | ⃑ 𝑗 + | | − one − 2 three 3 | | ⃑ π‘˜ = ( − 2 × 4 + 4 × iii ) ⃑ 𝑖 − ( − i × 4 + 4 × 3 ) ⃑ 𝑗 + ( − i × 3 + 2 × 3 ) ⃑ π‘˜ = 4 ⃑ 𝑖 − viii ⃑ 𝑗 + 3 ⃑ π‘˜ = ( iv , − eight , three ) .

Nosotros tin can determine the position vector ⃑ π‘Ÿ for a bespeak on the plane from either equation of the line, since both lines are contained on the airplane. For simplicity, nosotros tin substitute 𝑑 = 0 in the outset equation to determine the position vector of the signal as ⃑ π‘Ÿ = ( 1 , − i , − 3 ) .

Substituting the normal vector ( 4 , − viii , 3 ) and the position vector of a point on the plane ( 1 , − ane , − 3 ) , nosotros take ( 4 , − 8 , 3 ) ⋅ ⃑ π‘Ÿ = ( 4 , − eight , 3 ) ⋅ ( one , − i , − 3 ) = four × i + ( − 8 ) × ( − one ) + 3 × ( − 3 ) = three .

Thus, the equation of the airplane containing the two directly lines ⃑ π‘Ÿ and ⃑ π‘Ÿ in vector form is ( four , − eight , iii ) ⋅ ⃑ π‘Ÿ = 3 .

Key Points

  • The general form of the equation of a plane in ℝ is π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 + 𝑑 = 0 .
  • If ( π‘₯ , 𝑦 , 𝑧 ) is a point that lies on the plane, and so 𝑑 = − ( π‘Ž π‘₯ + 𝑏 𝑦 + 𝑐 𝑧 ) , and we can write the point–normal, or scalar, form of the equation of the airplane as π‘Ž ( π‘₯ − π‘₯ ) + 𝑏 ( 𝑦 − 𝑦 ) + 𝑐 ( 𝑧 − 𝑧 ) = 0 , where π‘Ž , 𝑏 , and 𝑐 are the components of the normal vector ⃑ 𝑛 = ( π‘Ž , 𝑏 , 𝑐 ) , which is perpendicular to the plane or any vector parallel to the plane.
  • If we are given a point on the plane ( π‘₯ , 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 airplane in vector grade can be written every bit ⃑ 𝑛 ⋅ ⃑ π‘Ÿ = ⃑ 𝑛 ⋅ ⃑ π‘Ÿ , with ⃑ π‘Ÿ = ( π‘₯ , 𝑦 , 𝑧 ) and ⃑ π‘Ÿ as the position vector of a point that lies on the plane.

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

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