Review Find an equation of the plane that passes through the point with a normal vector n
Kinh Nghiệm về Find an equation of the plane that passes through the point with a normal vector n Mới Nhất
Lê Hữu Kông đang tìm kiếm từ khóa Find an equation of the plane that passes through the point with a normal vector n được Cập Nhật vào lúc : 2022-11-26 16:44:03 . Với phương châm chia sẻ Bí quyết về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi tham khảo nội dung bài viết vẫn ko hiểu thì hoàn toàn có thể lại Comments ở cuối bài để Tác giả lý giải và hướng dẫn lại nha.$begingroup$
Nội dung chính Show- Definition: General Form of the Equation of a PlaneDefinition: Scalar Form of the Equation of a PlaneExample 1: Finding the Equation of a Plane given a Point and Its Normal VectorExample 2: Finding the General Equation of a Plane Passing through a Given Point and Parallel to Two Given VectorsExample 3: Finding the General Equation of a Plane Passing through Three Noncollinear PointsDefinition: Vector Form of the Equation of a PlaneExample 4: Finding the Vector Form of the Equation of a Plane given Its Normal Vector EquationExample 5: Finding the Vector Form of the Equation of a PlaneExample 6: Finding the Vector Equation of a Plane Containing Two Lines given Their Vector
Equations
The given points are $M(3,-1,2)$ and $M1(0,1,2)$ , plane A is passing though this 2 points. Plane B $2x-y+2z-1=0$ and its normal to A. How to we find the equation of plane A? I have tried with cross product of vector $MM1$ and vector $b=(2,-1,2)$, but it didn't work.
asked Aug 21, 2022 9:11
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I think it must work, when you do the cross product of $MM_1$ with $b$ you get $n=(-3,2,0)×(2,-1,2)=(4,6,-1)$.
This vector $n$ is a normal vector to plane A since the two vectors $b$ and $MM_1$ belong to plane $A$. So the equation of plane A is $4x+6y-z+c=0$.
Substitute the coordinates of $M$(or $M_1$) in the equation to get $c$. The equation is $4x+6y-z=4$.
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Perpendicular Planes
If you have a plane $P_1$ with equation $a_1 x + b_1 y + c_1 z = d_1$ and two points $p_1=(x_1, y_1, z_1)$ and $p_2=(x_2, y_2, z_2)$, then to find the plane $P_2$ with equation $a_2 x + b_2 y + c_2 z = d_2$ which is perpendicular to $P_1$ and contains both $p_1$ and $p_2$, you can use the following facts:
The vector $v_1:= (a_1, b_1, c_1)$ is perpendicular to $P_1$The vector $v_2:= (a_2, b_2, c_2)$ is perpendicular to $P_2$The fact that $P_1$ is perpendicular to $P_2$ implies that $v_1$ is perpendicular to $v_2$. The vector $v_3 := p_2 - p_1$ is parallel to $P_2$, and thus $v_3$ is perpendicular to $v_2$. The fact that $v_2$ is perpendicular to both $v_1$ and $v_3$ implies that $v_2$ is a multiple of the cross product of $v_1$ and $v_3$. You can safely set $v_2=v_1 times v_3$. (You should think about why this is true.) You can use fact 5 to give you acceptable values for $a_2$, $b_2$, and $c_2.$To figure out $d_2$, use the fact that $p_1 cdot v_2= d_2$ or $p_2 cdot v_2= d_2$.Fact 6 gave you $a_2$, $b_2$, and $c_2$. Fact 7 gave you $d_2$, so the problem is solved.answered Aug 21, 2022 10:56
irchansirchans
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In this explainer, we will learn 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.
Let’s first consider the equation of a line in Cartesian form and rewrite it in vector form in two 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 straight line in two dimensions is 𝑎𝑥+𝑏𝑦+ 𝑐=0.
This can also be written in the form 𝑦=𝑚𝑥+𝑑, where 𝑚 is the gradient and 𝑑 is the 𝑦-intercept, which we can determine by knowing two points on the line. If (𝑥,𝑦) is a point that lies on the line, we can determine 𝑐 from the general form as 𝑐=−(𝑎𝑥+𝑏𝑦); thus, the equation of the line can be written as 𝑎𝑥+𝑏𝑦−(𝑎𝑥+𝑏𝑦 )=0𝑎(𝑥−𝑥)+𝑏(𝑦−𝑦)=0.
The equation of the line can also be realized as a dot product of two vectors as (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 can be written in vector form as ⃑𝑛⋅⃑𝑟−⃑𝑟=0⃑𝑛⋅⃑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 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 ⃑𝑟−⃑𝑟 are perpendicular to each other by this property.
A normal vector ⃑𝑛 to a line or plane 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 we have ⃑𝑛⋅⃑𝑣=0, by the property of the dot product.
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 represent the equation of a plane as follows.
Definition: General Form of the Equation of a Plane
The general form of the equation of a plane in ℝ is 𝑎𝑥+𝑏𝑦+𝑐𝑧+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 plane.
If (𝑥,𝑦,𝑧) is a point that lies on the plane, then 𝑑=−(𝑎𝑥+𝑏𝑦+𝑐𝑧) and we can write the equation of the plane as 𝑎𝑥+𝑏𝑦+𝑐𝑧−(𝑎𝑥 +𝑏𝑦+𝑐𝑧)=0.
This can be rearranged to give the equation of the plane in scalar form.
Definition: Scalar Form of the Equation of a Plane
The scalar form of the equation of a plane in ℝ containing the point (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 example where we determine the equation of the plane in this form from the normal vector and a given point that lies on the plane.
Example 1: Finding the Equation of a Plane given a Point and Its Normal Vector
Give the equation of the plane with normal vector (10, 8,3) that contains the point (10,5,5).
Answer
In this example, we want to determine the equation of the plane by using one point on the plane and a given normal vector to the plane.
Recall that the scalar form of the equation of a plane with a normal vector ⃑𝑛=(𝑎,𝑏,𝑐) that contains the point (𝑥,𝑦,𝑧) is 𝑎(𝑥−𝑥)+𝑏(𝑦−𝑦)+𝑐(𝑧−𝑧)=0.
Thus, substituting the values for the given normal vector (10,8,3) and point (10,5,5), we have 10(𝑥−10)+8(𝑦−5)+3(𝑧−5)=010𝑥−100+8𝑦−40+3𝑧−15=010𝑥+8𝑦+3𝑧−155=0.
Thus, the general form of the equation of the plane with normal vector (10,8,3) that contains the point (10,5,5) is 10𝑥+8𝑦+3𝑧−155=0.
If we are given a point that lies on the plane, (𝑥,𝑦,𝑧), and two nonparallel vectors, ⃑𝑣 and ⃑𝑣, that are parallel to the plane, then we 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. Recall that the cross product of two vectors produces a vector that is perpendicular to both vectors. We 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 determine the equation of the plane by first finding the normal vector of the plane from two vectors that are parallel to it.
Example 2: Finding the General Equation of a Plane Passing through a Given Point and Parallel to Two Given Vectors
Find the general equation of the plane that passes through the point (5,1,−1) and is parallel to the two vectors (9,7,−8) and (−2,2,−1).
Answer
In this example, we want to determine 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=(𝑎,𝑏,𝑐) that contains the point (𝑥,𝑦,𝑧) is 𝑎(𝑥−𝑥)+𝑏( 𝑦−𝑦)+𝑐(𝑧−𝑧)=0 .
We need to determine 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 plane. We can find the normal vector by taking the cross product between these vectors: ⃑𝑛=( 9,7,−8)×(−2,2,−1)=||||⃑𝑖⃑𝑗⃑𝑘97−8−22−1| |||=||7−82−1| |⃑𝑖−||9−8−2−1||⃑𝑗+||97−22||⃑𝑘=(7×(−1)− (−8)×2)⃑𝑖−(9×(−1)−(−8)×(−2))⃑𝑗+(9×2−7×(−2))⃑𝑘=9⃑𝑖+25⃑𝑗+32⃑𝑘=(9,25,32).
Using the normal vector (9,25,32) and a point on the plane (5,1,−1), we have 9(𝑥−5)+25(𝑦−1)+32(𝑧+1)=09𝑥−45+25𝑦−25+32𝑧+32=09𝑥+25𝑦+32𝑧−38=0.
Thus, the general equation of the plane that passes through the point (5,1,−1) and is parallel to the two vectors (9,7,−8) and (−2,2,−1) is 9𝑥+25𝑦+32𝑧−38=0.
If a plane contains three points (𝑥,𝑦,𝑧), (𝑥,𝑦,𝑧), and (𝑥,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 we get 𝑎(𝑥−𝑥)+𝑏(𝑦−𝑦)+𝑐(𝑧−𝑧)=0,𝑎(𝑥−𝑥)+𝑏(𝑦−𝑦)+𝑐(𝑧−𝑧)=0, similar to how we can determine the equation of a line with two given points. However, this is not the standard way to determine the equation of a plane. Instead, we shall determine a normal vector by noting that the difference of the position vectors of any two points on the plane is a vector parallel to the plane; we shall revisit this when considering the vector form of the equation of the plane.
If we denote the position vectors of the three noncollinear points as ⃑𝑟=(𝑥,𝑦,𝑧) , ⃑𝑟=(𝑥,𝑦 ,𝑧), and ⃑𝑟=(𝑥,𝑦,𝑧) , then we can obtain two vectors parallel to the plane by subtracting pairs of these position vectors as ⃑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 order; for example, another choice could be ⃑𝑣=⃑𝑟−⃑𝑟,⃑𝑣=⃑𝑟−⃑𝑟.
In any case, the normal vector can be determined from the cross product of these two vectors: ⃑𝑛=⃑𝑣×⃑𝑣.
Now, let’s consider an example where we use this form along with information about three points that lie on the plane to determine the equation.
Example 3: Finding the General Equation of a Plane 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 lie on the plane.
Recall that the equation of a plane with a normal vector ⃑𝑛=(𝑎 ,𝑏,𝑐) that contains the point (𝑥,𝑦,𝑧) is 𝑎 (𝑥−𝑥)+𝑏(𝑦−𝑦)+ 𝑐(𝑧−𝑧)=0.
Let’s first determine a normal vector to the plane. We can obtain two vectors in the plane by subtracting the position vectors of pairs of points on the plane: ⃑𝑣=(1,0,3)−(1,2,−1)=(0,−2,4),⃑𝑣=(1,0,3)−(6,1,6)=(−5,−1,−3).
We can find the normal vector by taking the cross product between these vectors: ⃑𝑛=⃑𝑣×⃑𝑣=(0,−2,4)×(−5,−1,−3)=||||⃑𝑖⃑𝑗 ⃑𝑘0−24−5−1−3||||=||−24−1−3||⃑𝑖−||04−5−3||⃑𝑗+||0−2−5 −1||⃑𝑘=((−2)×(−3)−4×(−1))⃑𝑖−(0×(−3)−4×(−5))⃑𝑗+(0×(− 1)−(−2)×(−5)) ⃑𝑘=10⃑𝑖−20⃑𝑗−10⃑𝑘=(10,−20,−10).
Using the normal vector (10,−20,−10) and any of the given points that lie on the plane, for example, (1,0,3), the equation of the plane becomes 10(𝑥−1)−20(𝑦−0)−10(𝑧−3)=010𝑥−10−20𝑦−10𝑧+30=010𝑥−20𝑦−10𝑧+20=0.
Thus, dividing by 10, we obtain the equation of the plane in general form as 𝑥−2𝑦−𝑧+2=0.
The scalar equation of a plane can also be realized as the dot product of two 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 point on the plane and 𝑃=(𝑥,𝑦,𝑧) be any point 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 plane in the vector form can be written as ⃑𝑛⋅⃑𝑟−⃑𝑟=0.
The vector ⃑𝑛 is perpendicular to the plane, which means it is perpendicular to the vector of the difference of position vectors of any two points on the plane. This makes sense because, by construction, ⃑𝑟−⃑𝑟 will always 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 be rearranged to give the vector form 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 point 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, let’s look two examples where we determine the equation of the planes in vector form from given normal vectors and points that lie 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 plane that has normal vector ⃑𝑛 =⃑𝑖+⃑𝑗+⃑𝑘 and contains the point (2,6,6).
Answer
In this example, we want to determine the equation of a plane, in vector form, by using a point that lies on the plane and a given normal vector.
Recall that the vector equation of the plane can be written as ⃑𝑛⋅⃑𝑟=⃑𝑛⋅⃑𝑟, where ⃑𝑛 is a normal vector to the plane and ⃑𝑟 is the position vector of a point that lies on the plane.
The equation of the plane with normal vector ⃑𝑛=(1,1,1) that contains the point (2,6,6) with position vector ⃑𝑟=(2,6,6) is (1,1,1)⋅⃑𝑟=(1,1,1)⋅(2,6,6)=1×2+1×6+1×6=14.
Thus, the vector form of the equation of the plane is (1,1,1)⋅⃑𝑟=14.
Now, let’s consider an example where we convert the equation of the plane from general form to vector form.
Example 5: Finding the Vector Form of the Equation of a Plane
The equation of a plane has the general form 5𝑥+6𝑦+9𝑧−28=0. What is its vector form?
Answer
In this example, we want to determine the equation of the plane in vector form by using the given equation of the plane in general form.
Recall that the general form of the equation of a plane in ℝ is 𝑎𝑥+𝑏𝑦+𝑐𝑧 +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 plane. The vector equation of the plane can be written as ⃑𝑛⋅⃑𝑟=−𝑑.
From the given equation of the plane, 5 𝑥+6𝑦+9𝑧−28=0, we can identify the normal vector as ⃑𝑛=(5,6,9 ) and 𝑑=−28. The vector equation of the plane can be written as (5,6,9)⋅⃑𝑟=28.
The equation of a straight line in ℝ in vector form is ⃑𝑟=⃑𝑟+𝑡⃑𝑣,𝑡∈ℝ,line where ⃑𝑟=(𝑥,𝑦,𝑧) is a position vector of a point 𝑃=(𝑥,𝑦,𝑧) on the line and ⃑𝑣= (𝑎,𝑏,𝑐) is a vector parallel to the line.
If we have a plane containing two district intersecting straight lines with vector equations ⃑𝑟=⃑𝑎+𝑡⃑𝑣,⃑𝑟=⃑𝑎+𝑡⃑𝑣 , then we can determine a point that lies on the plane from either of these equations. For simplicity, we can substitute 𝑡=0 in the first equation, giving the position vector of a point 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 need to determine 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 can obtain the normal vector to the plane by taking the cross product of the two vectors: ⃑𝑛=⃑𝑣×⃑𝑣.
By putting these together, the equation of the plane can be written as ⃑𝑣×⃑𝑣⋅⃑𝑟=⃑𝑣×⃑𝑣⋅⃑𝑎.
In our final example, we will determine the equation of a plane in vector form from the vector equations of two straight lines that lie on the plane.
Example 6: 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 two straight lines ⃑𝑟=⃑𝑖−⃑𝑗−3⃑𝑘+𝑡3⃑𝑖+3⃑𝑗+4⃑𝑘 and ⃑𝑟=−⃑𝑖−2⃑𝑗−3⃑𝑘+𝑡−⃑𝑖−2⃑𝑗−4⃑𝑘.
Answer
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 form of the equation of the plane can be written as ⃑𝑛⋅⃑𝑟=⃑𝑛⋅⃑𝑟, where ⃑𝑛 is a normal vector to the plane and ⃑𝑟 is the position vector of a point that lies on the plane.
For simplicity, let’s begin by writing the vector equations of the straight lines as ⃑𝑟=(1, −1,−3)+𝑡(3,3,4),⃑𝑟=(−1,−2, −3)+𝑡(−1,−2,−4).
We note that, from the vector equations of the straight lines, the vector (3,3,4) is parallel to the first line and (−1,−2,−4) is parallel to the second, which means both are parallel to the plane. Thus, in order to determine a normal vector, ⃑𝑛, to the plane, we need to find a vector that is perpendicular to both (−1,−2,−4) and (3,3,4). We can do this by taking the cross product: ⃑𝑛=(−1,−2,−4)×(3,3,4)=||||⃑𝑖⃑𝑗⃑𝑘− 1−2−4334||||=||−2−434||⃑𝑖− ||−1−434||⃑𝑗+||−1−233||⃑𝑘=(−2×4+4×3)⃑𝑖 −(−1×4+4×3)⃑𝑗+(−1×3+2×3)⃑𝑘=4⃑𝑖−8⃑𝑗+3⃑ 𝑘=(4,−8,3).
We 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 ( 4,−8,3) and the position vector of a point on the plane (1,−1,−3), we have (4,−8,3)⋅⃑𝑟=( 4,−8,3)⋅(1,−1,−3)=4×1+(−8)×(−1)+3×(−3)=3 .
Thus, the equation of the plane containing the two straight lines ⃑𝑟 and ⃑𝑟 in vector form is (4,−8,3)⋅⃑𝑟=3.
Key Points
- The general form of the equation of a plane in ℝ is 𝑎𝑥+𝑏𝑦+𝑐𝑧+
𝑑=0.If (𝑥,𝑦,𝑧) is a point that lies on the plane, then 𝑑
=−(𝑎𝑥+𝑏𝑦+𝑐𝑧), and we can write the point–normal, or scalar, form of the equation of the plane 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 plane or any vector parallel to the plane.If we are given a point on the plane (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 can be written as ⃑𝑛⋅⃑𝑟=⃑𝑛⋅⃑𝑟, with ⃑𝑟=(𝑥,𝑦,𝑧) and ⃑𝑟 as the position vector of a point that lies on the plane.
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