Hello everyone, yesterday I guided you how to write equation of circle and how to write equation of a line in space already.

So today I will continue to show you how to write equations of planes in space. Specifically, write the equation of the plane when it is known:

- A passing point and a normal vector
- Three passing points

Corresponding to each case, I will start with a presentation of the theory and end with one or several corresponding illustrative examples:

**Case 1. Knowing a passing point and a normal vector**

The plane equation passes through the point $M_0=(x_0; y_0; z_0)$ and takes the vector $\vec{n}=(A; B; C)$ which is different from the vector $\vec{0}=(0; 0; 0)$ as the normal vector will be $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$

**Example 1.** Write an equation for the plane $(P)$ knowing the plane $(P)$ passing through the point $M=(2; 3; 5)$ and getting the vector $\vec{n}=(7; 11; 13) $ as normal vector

**The answer:**

The plane equation $(P)$ has the form $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$

Since $(P)$ passes through $M=(2; 3; 5)$ and takes the vector $\vec{n}=(7; 11; 13)$ as the normal vector, the equation of the plane $ (P)$ is $7(x-2)+11(y-3)+13(z-5)=0$ or $7x+11y+13z-112=0$

So => equation of plane $(P)$ is $7x+11y+13z-112=0$

**Case #2. Know the three passing points**

**Method 1.**Suitable for high school students, college and university students**Way 2.**Only suitable for College and University students

Please note that

Way 2not suitable for high school students because they have not learned the definition of a matrix and of course cannot calculate the determinant of a square matrix of level 3 x 3.

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*

**Method 1. Based on the directed product of two vectors**

Suppose the plane $(P)$ passes through three points $A=(x_a; y_a; z_a); B=(x_b; y_b; z_b); C=(x_c; y_c; z_c)$

We can then find the equation of the plane $(P)$ by the steps below:

**Step 1.** Calculate vector $\overrightarrow{AB}=(x_b-x_a; y_b-y_a; z_b-z_a), \overrightarrow{AC}=(x_c-x_a; y_c-y_a; z_c-z_a)$

**Step 2.** Calculate the product of the direction of the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

Here I will put $\overrightarrow{AB}=(x_1; y_1; z_1), \overrightarrow{AC}=(x_2; y_2; z_2)$ for convenience of presentation

$[\overrightarrow{AB}; \overrightarrow{AC}]=(x_2y_3-y_2x_3; x_3y_1-y_3x_1; x_1y_2-y_1x_2)$

The formula is quite long, but that’s okay because we can calculate it with a calculator CASIO fx-580VN X

**Attention:**

- The directed product of two vectors is a vector.
- The directed product of the vectors $\vec{a}$ and $\vec{b}$ is usually denoted by $[\vec{a}; \vec{b}]$ or $\vec{a} \land \vec{b}$.

Then the vector $\vec{n}=[\overrightarrow{AB}; \overrightarrow{AC}]$ is a normal vector to the plane $(P)$

**Step 3.** Write the equation of the plane passing through the point A or B or C and take the vector $\vec{n}$ as the normal vector.

**Method 2. Based on the determinant of the square matrix of order 3 x 3**

Given three noncollinear points $A=(x_a; y_a; z_a); B=(x_b; y_b; z_b); C=(x_c; y_c; z_c)$ the equation of the plane $(P)$ passing through the three points just given is

$\left|\begin{array}{ccc} x-x_{a} & y-y_{a} & z-z_{a}\\ x_{b}-x_{a} & y_{b}-y_ {a} & z_{b}-z_{a}\\ x_{c}-x_{a} & y_{c}-y_{a} & z_{c}-z_{a}\end{array}\ right|=0$

**Example 2.** In the space $Oxyz$ for three points $A=(2; -1; 3), B=(4; 0; 1), C=(-10; 5; 3)$. Write the equation of the plane $(P)$ passing through the three points just given

**The answer:**

**Method 1. Based on the directed product of two vectors**

$\overrightarrow{AB}=(2; 1; -2), \overrightarrow{AC}=(-12; 6; 0)$

The directed product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is $\vec{n}=[\overrightarrow{AB}; \overrightarrow{AC}]=(12; 24; 24)$

Since $(P)$ passes through $A=(2; -1; 3)$ and takes the vector $\vec{n}=(12; 24; 24)$ as the normal vector, the equation of the plane $(P)$ is $12(x-2)+24(y+1)+24(z-3)=0$ or $x+2y+2z-6=0$

So => equation of plane $(P)$ is $x+2y+2z-6=0$

**Method 2. Based on the determinant of the square matrix of order 3 x 3**

The plane equation $(P)$ is $\left|\begin{array}{ccc} x-2 & y+1 & z-3\\ 4-2 & 0+1 & 1-3\\ -10 -2 & 5+1 & 3-3\end{array}\right|=0$

$\Leftrightarrow \left|\begin{array}{ccc} x-2 & y+1 & z-3\\ 2 & 1 & -2\\ -12 & 6 & 0\end{array}\right|= 0 \Leftrightarrow 12x-72+24z+24y=0 \Leftrightarrow x+2y+2z-6=0$

So the equation of the plane $(P)$ is $x+2y+2z-6=0$

**Example 3.** In the space $Oxyz$ for three points $A=(1; 0; 0), B=(0; 2; 0), C=(0; 0; 3)$. Write the equation of the plane $(P)$ passing through the three points just given

**Comment:**

The coordinates of the three points that the plane passes through are of the form $(a; 0; 0), (0; b; 0), (0; 0; c)$, where a, b, c are other real numbers, respectively. 0

This is a special case, we should use a special way to save time and effort to solve.

Specifically, the equation of the plane passing through three points $(a; 0; 0), (0; b; 0), (0; 0; c)$ where a, b, c are non-zero real numbers $\ frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

**The answer:**

Applying the equation of the plane along the intercept we get the equation where the plane $(P)$ is $\frac{x}{1}+\frac{y}{2}+\frac{z}{3} =1$ or $6x+3y+2z-6=0$

Attention:Of course, you can still useWay 1orWay 2to find the equation of this plane.

**Trick to quickly calculate the directed product of two vectors**

This trick is quite a lot of operations but it is not complicated, after understanding and getting used to your hand, it only takes less than a minute to complete and find the answer.

Here I will illustrate $\overrightarrow{AB}=(2; 1; -2)$ and $\overrightarrow{AC}=(-12; 6; 0)$

**Step 1.** Choose the method to calculate Vector

**Step 2.** Assign the vector $\overrightarrow{AB}$ to `VctA`

**Step 2.1.** Choose `VctA`

**Step 2.2.** Press the number key `3`

**Step 2.3.** Enter coordinates for vector `VctA`

**Step 3.** Assign the vector $\overrightarrow{AC}$ to `VctB`

**Step 3.1.** Press `OPTN`

=> choose `Define Vector`

**Step 3.2.** Choose `VctB`

**Step 3.3.** Press the number key `3`

**Step 3.4.** Enter coordinates for vector `VctB`

**Step 4.** Directional product of `VctA`

and `VctB`

**Step 4.1.** Press `AC`

=> press `OPTN`

=> press number key `3`

**Step 4.2.** Press `“nhân”`

**Step 4.3.** Press `OPTN`

=> press number key `4`

**Step 4.4.** Press `=`

So => $[\overrightarrow{AB}; \overrightarrow{AC}]=(12; 24; 24)$

**Epilogue**

Yes, here are 2 cases **write equation of plane** most commonly encountered.

Between **Case 1** and **Case 2** there is one more intermediate case for writing the plane equation when **know a passing point and two direction vectors.**

If this is the case, you can refer to one of the two cases above and then proceed to write by relying on the existing knowledge.

Hope this article will be useful to you. Goodbye and see you in the next posts!

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