Instruction

1

Let the plane specified by three points belonging to it K(xk;yk;zk), M(xm;ym;zm), P(xp;yp;zp). To find the normal vector to make the equation of this

**plane**. Mark an arbitrary point lying on**the plane**, with the letter L, I want her to have coordinates (x;y;z). Now consider the three vectors PK, PM and PL, they lie on the same**plane**(coplanar), so their scalar triple product is zero.2

Find the coordinates of vectors PK, PM, and PL:

PK = (xk-xp, yk-yp;zk-zp)

PM = (xm-xp, ym-yp;zm-zp)

PL = (x-xp, y-yp z-zp)

Mixed product of these vectors is equal to the determinant presented in the figure. This determinant should be calculated to find the equation for

PK = (xk-xp, yk-yp;zk-zp)

PM = (xm-xp, ym-yp;zm-zp)

PL = (x-xp, y-yp z-zp)

Mixed product of these vectors is equal to the determinant presented in the figure. This determinant should be calculated to find the equation for

**the plane**. The calculation of the mixed product for a particular case, see example.3

Example

Let a plane defined by three points K(2;1;-2), M(0;0;-1) and P(1;8;1). You want to find a normal vector

Take an arbitrary point L with coordinates (x;y;z). Compute the vectors PK, PM, and PL:

PK = (2-1;1-8;-2-1) = (1;-7;-3)

PM = (0-1;0-8;-1-1) = (-1;-8;-2)

PL = (x-1;y-8;z-1)

Make the determinant for the scalar triple product of vectors (it is on the picture).

Let a plane defined by three points K(2;1;-2), M(0;0;-1) and P(1;8;1). You want to find a normal vector

**to the plane**.Take an arbitrary point L with coordinates (x;y;z). Compute the vectors PK, PM, and PL:

PK = (2-1;1-8;-2-1) = (1;-7;-3)

PM = (0-1;0-8;-1-1) = (-1;-8;-2)

PL = (x-1;y-8;z-1)

Make the determinant for the scalar triple product of vectors (it is on the picture).

4

Now decompose the determinant along the first row, and then count the value of the determinant of size 2 by 2.

Thus, the equation

Thus, the equation

**of plane**is 10x + 5y - 15z - 15 = 0 or what is the same, -2x + y - 3z - 3 = 0. It is easy to determine the vector normal to**plane**: n = (-2;1;-3).# Advice 2 : How to calculate the vector

Vector as a directed line segment, does not only depend on the absolute value (modulus), which is equal to its length. Another important characteristic – direction vector. It can be defined as coordinates, and the angle between the vector and the coordinate axis. Vector calculation is also produced when finding the sum and difference of vectors.

You will need

- - definition vector;
- properties of vectors;
- calculator;
- - table Bradis or PC.

Instruction

1

To calculate a vector, knowing its coordinates. To do this, define the coordinates of the beginning and end of the vector. Let them be equal to (x1;y1) and (x2;y2). To make the calculation of a vector, find its coordinates. To do this, from the coordinates of end of vector, subtract the coordinates of its beginning. They will be equal to (x2 - x1;y2-y1). Take x= x2 - x1; y= y2-y1, then the coordinates of the vector will equal (x;y).

2

Determine the length of the vector. This can be done simply by measuring it with a ruler. But if you know the coordinates of the vector, calculate length. To do this, find the sum of squares of coordinates of the vector, and extract the resulting number square root. Then the length of the vector is equal to d=√(x2+y2).

3

Then find the direction vector. To do this, determine the angle α between it and axis OX. The tangent of that angle is equal to the y coordinate of the vector to the x-coordinate of the (tg α= y/x). To find the angle, use the calculator inverse tangent function, table Bradis or PC. Knowing the length of the vector and its direction relative to the axis, it is possible to find the position in space of any vector.

4

Example:

the coordinates of start vector is (-3;5) and the coordinates of the end (1;7). Find the coordinates of the vector (1-(-3);7-5)=(4;2). Then its length will be d=√(42+22)=√20≈4,47 linear units. The tangent of the angle between the vector and the axis OX will be tg α=2/4=0,5. The tangent of this angle is roughly equal to 26.6°.

the coordinates of start vector is (-3;5) and the coordinates of the end (1;7). Find the coordinates of the vector (1-(-3);7-5)=(4;2). Then its length will be d=√(42+22)=√20≈4,47 linear units. The tangent of the angle between the vector and the axis OX will be tg α=2/4=0,5. The tangent of this angle is roughly equal to 26.6°.

5

Find a vector that is the sum of two vectors whose coordinates are known. You should put the corresponding coordinates of the vectors that add up. If the coordinates of the vectors, which are formed, are respectively(x1;y1) and (x2;y2), their sum will be equal to the vector with coordinates ((x1+x2;y1+y2)). If you want to find the difference of two vectors, find the sum, pre-multiplying the coordinates of the vector which is subtracted by -1.

6

If you know the lengths of the vectors d1 and d2, and the angle between them α, find the sum, using the theorem of cosines. To do this, find the sum of the squares of the lengths of the vectors and from the resulting number, subtract twice the product of these lengths multiplied by the cosine of the angle between them. From the numbers, extract the square root. This will be the length of the vector that is the sum of these two vectors (d=√(d12+d22-d1∙d2∙Cos(α)).

# Advice 3 : How to find the normal vector

The task of finding

**the vector****of the normal**straight line on the plane and the plane in space is too easy. In fact, it ends with the entry of General equations of straight line or plane. Because the curve on the plane just a special case of a surface in space, it is about the normals to the surface will be discussed.Instruction

1

The first method This method is the easiest, but its understanding requires knowledge of the concept of the scalar field. However, inexperienced in this matter, the reader will be able to use the resulting formulas of this issue.

2

It is known that the scalar field f is defined as f=f(x, y, z) and any surface is the level surface f(x, y, z)=C (C=const). In addition, the normal surface level coincides with the gradient of the scalar field at a given point.

3

The gradient of the scalar field (function of three variables) is called a vector g=gradf=iдf/DH+jдf/DN+kдf/дz={дf/HH, дf/du, дf/дz}. Since the length

**of the normals**is irrelevant, it remains only to record the answer. Normal to поверхностиf(x, y, z)-C=0 in точкеМ0(x0, y0, z0) n=gradf=iдf/DH+jдf/DN+kдf/дz={дf/HH, дf/du, дf/дz}.4

*The second method Let surface given by the equation F(x, y, z)=0. To be able to continue to draw analogies with the first method, note that the derivative of a constant is zero and F is given as f(x, y, z)-C=0 (C=const). If you hold a section of this surface by an arbitrary plane, the resulting spatial curve can be considered as the locus of any vector function r(t)= ix(t)x+jy(t)+kz(t). Then the derivative*

**vector**r’(t)= ix’(t)+jy’(t)+kz’(t) is tangential at a point M0(x0, y0, z0) of the surface (see Fig.1).

5

In order to avoid confusion, the current coordinates of the tangent line should be marked, e.g. in italics (x, y, z). The canonical equation of the tangent line, given that r’(t0) is the direction vector, is written as (x-x(t0))/(dx(t0)/dt)= (y-y(t0))/(dy(t0)/dt)= (z-z(t0))/(dz(t0)/dt).

6

Substituting the coordinates of the vector-function in the equation of the surface f(x, y, z)-C=0 and differentiated with respect to t you will receive (дf/DH)(DH/дt)+(дf/du) (du/дt)+(дf/дz)(дz/дt)=0. The equality is a scalar product of some

**vector**n(дf/HH, дf/du, дf/дz) and r’(x’(t), y’(t), z’(t)). Since it is zero, then n(дf/HH, дf/du, дf/дz) is the desired vector**normals**. It is obvious that the results of both methods are identical.7

Example (has a theoretical value). To find a vector

**normal**to the surface given by the classical equation of two variables function z=z(x, y). Solution. Rewrite this equation in the form z-z(x, y)=F(x, y, z)=0. Following any of the suggested methods, it turns out that n(-дz/DX, -дz/dy, 1) is the sought vector**of the normal**.# Advice 4 : How to sort vector

Any

**vector**can be decomposed into the sum of**vector**s, and such possibilities are endless. The task to decompose**the vector**can be given as in the geometric form, and the form of formulas, this will depend on the solution of the problem.You will need

- - initial vector;
- - vector for which you want to decompose.

Instruction

1

If you want to decompose

**a vector**in the drawing, select the direction for parts. For convenience of calculations most often used decomposition of the**vector**a, parallel to the coordinate axes, but you can choose absolutely any convenient direction.2

Draw one of the components of

**the vector**s, in which case it must proceed from the same point as the original (the length you choose). Connect the ends of the source and the received**vector**and another**vector**om. Please note: two of the received**vector**and the result should lead you to the same point as the source (if you move the arrows).3

Move the resulting

**vector**and in the place where they are convenient to use, while maintaining the direction and length. Regardless of where**the vector**and will be, in total they will be equal to the original. Please note that if you place the resulting**vector**and so that they proceeded from the same point as the source, and the dotted line connecting their ends will produce a parallelogram, and the original**vector**will coincide with one of the diagonals.4

If you need to decompose

**the vector**{x1,x2,X3} with respect to the basis, that is, given**the vector**of am {P1, P2, P3}, {q1,q2,q3}, {r1,r2,r3}, we proceed as follows. Substitute the coordinate values into the formula x=ar+βq+γr.5

As a result, you will obtain a system of three equations P1A+q1β+r1γ=x1, p2α+q2β+r2γ=x2, p3α+q3β+r3γ=X3. Solve this system using method of summations or matrices, find the coefficients α, β, γ. If the task is given in the plane, the solution is more simple, because instead of three variables and equations you will receive only two (they will be in the form P1A+q1β=x1, p2α+q2β=x2). Write your answer in the form x=AP+βq+γr.

6

If you get an infinite number of solutions, make a conclusion that

**the vector**s p, q, r lie in the same plane with**the vector**ω of x and decompose it in the specified way definitely not.7

If solutions, the system has to write the answer to the problem:

**the vector**s p, q, r lie in one plane, and**the vector**x in another, so it is impossible to decompose the given image.