This article reveals the meaning of the perpendicularity of two vectors on a plane in three-dimensional space and finding the coordinates of a vector perpendicular to one or a whole pair of vectors. The topic is applicable to problems involving equations of lines and planes.
We will consider the necessary and sufficient condition for the perpendicularity of two vectors, solve the method of finding a vector perpendicular to a given one, and touch upon situations of finding a vector that is perpendicular to two vectors.
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Necessary and sufficient condition for the perpendicularity of two vectors
Let's apply the rule about perpendicular vectors on the plane and in three-dimensional space.
Definition 1
Provided the angle between two non-zero vectors is equal to 90 ° (π 2 radians) is called perpendicular.
What does this mean, and in what situations is it necessary to know about their perpendicularity?
Establishing perpendicularity is possible through the drawing. When plotting a vector on a plane from given points, you can geometrically measure the angle between them. Even if the perpendicularity of the vectors is established, it will not be entirely accurate. Most often, these tasks do not allow you to do this using a protractor, so this method is applicable only when nothing else is known about the vectors.
Most cases of proving the perpendicularity of two non-zero vectors on a plane or in space are done using necessary and sufficient condition for the perpendicularity of two vectors.
Theorem 1
The scalar product of two non-zero vectors a → and b → equal to zero to satisfy the equality a → , b → = 0 is sufficient for their perpendicularity.
Evidence 1
Let the given vectors a → and b → be perpendicular, then we will prove the equality a ⇀ , b → = 0 .
From the definition of dot product of vectors we know that it equals the product of the lengths of given vectors and the cosine of the angle between them. By condition, a → and b → are perpendicular, which means, based on the definition, the angle between them is 90 °. Then we have a → , b → = a → · b → · cos (a → , b → ^) = a → · b → · cos 90 ° = 0 .
Second part of the proof
Provided that a ⇀, b → = 0, prove the perpendicularity of a → and b →.
In fact, the proof is the opposite of the previous one. It is known that a → and b → are non-zero, which means that from the equality a ⇀ , b → = a → · b → · cos (a → , b →) ^ we find the cosine. Then we get cos (a → , b →) ^ = (a → , b →) a → · b → = 0 a → · b → = 0 . Since the cosine is zero, we can conclude that the angle a →, b → ^ of the vectors a → and b → is equal to 90 °. By definition, this is a necessary and sufficient property.
Perpendicularity condition on the coordinate plane
Chapter scalar product in coordinates demonstrates the inequality (a → , b →) = a x · b x + a y · b y , valid for vectors with coordinates a → = (a x , a y) and b → = (b x , b y), on the plane and (a → , b → ) = a x · b x + a y · b y for vectors a → = (a x , a y , a z) and b → = (b x , b y , b z) in space. The necessary and sufficient condition for the perpendicularity of two vectors in the coordinate plane is a x · b x + a y · b y = 0, for three-dimensional space a x · b x + a y · b y + a z · b z = 0.
Let's put it into practice and look at examples.
Example 1
Check the property of perpendicularity of two vectors a → = (2, - 3), b → = (- 6, - 4).
Solution
To solve this problem, you need to find the scalar product. If according to the condition it is equal to zero, then they are perpendicular.
(a → , b →) = a x · b x + a y · b y = 2 · (- 6) + (- 3) · (- 4) = 0 . The condition is met, which means that the given vectors are perpendicular to the plane.
Answer: yes, the given vectors a → and b → are perpendicular.
Example 2
Coordinate vectors i → , j → , k → are given. Check whether the vectors i → - j → and i → + 2 · j → + 2 · k → can be perpendicular.
Solution
In order to remember how vector coordinates are determined, you need to read the article about vector coordinates in a rectangular coordinate system. Thus, we find that the given vectors i → - j → and i → + 2 · j → + 2 · k → have corresponding coordinates (1, - 1, 0) and (1, 2, 2). We substitute the numerical values and get: i → + 2 · j → + 2 · k → , i → - j → = 1 · 1 + (- 1) · 2 + 0 · 2 = - 1 .
The expression is not equal to zero, (i → + 2 j → + 2 k →, i → - j →) ≠ 0, which means that the vectors i → - j → and i → + 2 j → + 2 k → are not perpendicular, since the condition is not met.
Answer: no, the vectors i → - j → and i → + 2 · j → + 2 · k → are not perpendicular.
Example 3
Given vectors a → = (1, 0, - 2) and b → = (λ, 5, 1). Find the value of λ at which these vectors are perpendicular.
Solution
We use the condition of perpendicularity of two vectors in space in square form, then we get
a x b x + a y b y + a z b z = 0 ⇔ 1 λ + 0 5 + (- 2) 1 = 0 ⇔ λ = 2
Answer: the vectors are perpendicular at the value λ = 2.
There are cases when the question of perpendicularity is impossible even under a necessary and sufficient condition. Given the known data on the three sides of a triangle on two vectors, it is possible to find angle between vectors and check it out.
Example 4
Given a triangle A B C with sides A B = 8, A C = 6, B C = 10 cm. Check the vectors A B → and A C → for perpendicularity.
Solution
If the vectors A B → and A C → are perpendicular, the triangle A B C is considered rectangular. Then we apply the Pythagorean theorem, where B C is the hypotenuse of the triangle. The equality B C 2 = A B 2 + A C 2 must be true. It follows that 10 2 = 8 2 + 6 2 ⇔ 100 = 100. This means that A B and A C are legs of the triangle A B C, therefore, A B → and A C → are perpendicular.
It is important to learn how to find the coordinates of a vector perpendicular to a given one. This is possible both on the plane and in space, provided that the vectors are perpendicular.
Finding a vector perpendicular to a given one in a plane.
A non-zero vector a → can have an infinite number of perpendicular vectors on the plane. Let's depict this on the coordinate line.
Given a non-zero vector a → lying on the straight line a. Then a given b →, located on any line perpendicular to line a, becomes perpendicular to a →. If the vector i → is perpendicular to the vector j → or any of the vectors λ · j → with λ equal to any real number other than zero, then finding the coordinates of the vector b → perpendicular to a → = (a x , a y) is reduced to an infinite set of solutions. But it is necessary to find the coordinates of the vector perpendicular to a → = (a x , a y) . To do this, it is necessary to write down the condition of perpendicularity of vectors in the following form: a x · b x + a y · b y = 0. We have b x and b y, which are the desired coordinates of the perpendicular vector. When a x ≠ 0, the value of b y is non-zero, and b x can be calculated from the inequality a x · b x + a y · b y = 0 ⇔ b x = - a y · b y a x. For a x = 0 and a y ≠ 0, we assign b x any value other than zero, and find b y from the expression b y = - a x · b x a y .
Example 5
Given a vector with coordinates a → = (- 2 , 2) . Find a vector perpendicular to this.
Solution
Let us denote the desired vector as b → (b x , b y) . Its coordinates can be found from the condition that the vectors a → and b → are perpendicular. Then we get: (a → , b →) = a x · b x + a y · b y = - 2 · b x + 2 · b y = 0 . Let's assign b y = 1 and substitute: - 2 · b x + 2 · b y = 0 ⇔ - 2 · b x + 2 = 0 . Hence, from the formula we get b x = - 2 - 2 = 1 2. This means that the vector b → = (1 2 , 1) is a vector perpendicular to a → .
Answer: b → = (1 2 , 1) .
If the question is raised about three-dimensional space, the problem is solved according to the same principle. For a given vector a → = (a x , a y , a z) there is an infinite number of perpendicular vectors. Will fix this on a three-dimensional coordinate plane. Given a → lying on the line a. The plane perpendicular to straight a is denoted by α. In this case, any non-zero vector b → from the plane α is perpendicular to a →.
It is necessary to find the coordinates of b → perpendicular to the non-zero vector a → = (a x , a y , a z) .
Let b → be given with coordinates b x , b y and b z . To find them, it is necessary to apply the definition of the condition of perpendicularity of two vectors. The equality a x · b x + a y · b y + a z · b z = 0 must be satisfied. From the condition a → is non-zero, which means that one of the coordinates has a value not equal to zero. Let's assume that a x ≠ 0, (a y ≠ 0 or a z ≠ 0). Therefore, we have the right to divide the entire inequality a x · b x + a y · b y + a z · b z = 0 by this coordinate, we obtain the expression b x + a y · b y + a z · b z a x = 0 ⇔ b x = - a y · b y + a z · b z a x . We assign any value to the coordinates b y and b x, calculate the value of b x based on the formula, b x = - a y · b y + a z · b z a x. The desired perpendicular vector will have the value a → = (a x, a y, a z).
Let's look at the proof using an example.
Example 6
Given a vector with coordinates a → = (1, 2, 3) . Find a vector perpendicular to the given one.
Solution
Let us denote the desired vector by b → = (b x , b y , b z) . Based on the condition that the vectors are perpendicular, the scalar product must be equal to zero.
a ⇀ , b ⇀ = 0 ⇔ a x b x + a y b y + a z b z = 0 ⇔ 1 b x + 2 b y + 3 b z = 0 ⇔ b x = - (2 b y + 3 b z)
If the value of b y = 1, b z = 1, then b x = - 2 b y - 3 b z = - (2 1 + 3 1) = - 5. It follows that the coordinates of the vector b → (- 5 , 1 , 1) . Vector b → is one of the vectors perpendicular to the given one.
Answer: b → = (- 5 , 1 , 1) .
Finding the coordinates of a vector perpendicular to two given vectors
We need to find the coordinates of the vector in three-dimensional space. It is perpendicular to the non-collinear vectors a → (a x , a y , a z) and b → = (b x , b y , b z) . Provided that the vectors a → and b → are collinear, it will be sufficient to find a vector perpendicular to a → or b → in the problem.
When solving, the concept of a vector product of vectors is used.
Vector product of vectors a → and b → is a vector that is simultaneously perpendicular to both a → and b →. To solve this problem, the vector product a → × b → is used. For three-dimensional space it has the form a → × b → = a → j → k → a x a y a z b x b y b z
Let's look at the vector product in more detail using an example problem.
Example 7
The vectors b → = (0, 2, 3) and a → = (2, 1, 0) are given. Find the coordinates of any vector perpendicular to the data simultaneously.
Solution
To solve, you need to find the vector product of vectors. (Please refer to paragraph calculating the determinant of a matrix to find the vector). We get:
a → × b → = i → j → k → 2 1 0 0 2 3 = i → 1 3 + j → 0 0 + k → 2 2 - k → 1 0 - j → 2 3 - i → 0 2 = 3 i → + (- 6) j → + 4 k →
Answer: (3 , - 6 , 4) - coordinates of a vector that is simultaneously perpendicular to the given a → and b → .
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Condition for vectors to be perpendicular
Vectors are perpendicular if and only if their dot product is zero.
Given two vectors a(xa;ya) and b(xb;yb). These vectors will be perpendicular if the expression xaxb + yayb = 0.
Vectors are parallel if their cross product is zero
Equation of a straight line on a plane. Basic problems on a straight line on a plane.
Any straight line on the plane can be specified by a first-order equation Ax + By + C = 0, and the constants A and B are not equal to zero at the same time, i.e. A2 + B2 0. This first-order equation is called the general equation of the line. Depending on the values of the constants A, B and C, the following special cases are possible: - C = 0, A 0, B 0 – the straight line passes through the origin - A = 0, B 0, C 0 ( By
C = 0) - straight line parallel to the Oy axis - B = 0, A 0, C 0 ( Ax + C = 0) - straight line parallel to the Oy axis - B = C = 0, A 0 - straight line coincides with the Oy axis - A = C = 0, B 0 – the straight line coincides with the Ox axis. The equation of the straight line can be presented in different forms depending on any given initial conditions.
If at least one of the coefficients A, B, C of level Ax+By+C=0 is equal to 0, level
called incomplete. By the form of the equation of a straight line one can judge its position on
flatness OXU. Possible cases:
1 C=0 L: Ax+By=0 t. O(0,0) satisfies this equation, which means it’s straight
passes through the origin
2 A=0 L: Ву+С=0 - normal rotation n=(0,B) is perpendicular to the OX axis from here
it follows that the straight line is parallel to the OX axis
3 V = 0 L: Ay+C=0 0 - nominal value n=(A,0) is perpendicular to the OY axis from here
it follows that the straight line is parallel to the axis of the op-amp
4 A=0, C=0 L: By=0(y=0(L=OX
5 B=0, C=0 L: Ax=0(x=0(L=OY
6 A (0, B (0, C (0 L; - does not pass through the origin and intersects
both axes.
Equation of a straight line on a plane passing through two given points and: 
Angle between planes.
Calculation of determinants
The calculation of determinants is based on their known properties, which apply to determinants of all orders. These are the properties:
1. If you rearrange two rows (or two columns) of the determinant, the determinant will change sign.
2. If the corresponding elements of two columns (or two rows) of the determinant are equal or proportional, then the determinant is equal to zero.
3. The value of the determinant will not change if you swap the rows and columns, maintaining their order.
4. If all the elements of a row (or column) have a common factor, then it can be taken out of the determinant sign.
5. The value of the determinant will not change if the corresponding elements of another row (or column) are added to the elements of one row (or column), multiplied by the same number.
The Matrix and the actions above them
Matrix- a mathematical object written in the form of a rectangular table of numbers (or elements of a ring) and allowing algebraic operations (addition, subtraction, multiplication, etc.) between it and other similar objects. Typically, matrices are represented as two-dimensional (rectangular) tables. Sometimes multidimensional matrices or non-rectangular matrices are considered.
Typically, the matrix is denoted by a capital letter of the Latin alphabet and highlighted with round brackets “(…)” (also marked with square brackets “[…]” or double straight lines “||…||”).
The numbers that make up the matrix (matrix elements) are often denoted by the same letter as the matrix itself, but lowercase (for example, a11 is an element of matrix A).
Each matrix element has 2 subscripts (aij) - the first “i” denotes the row number in which the element is located, and the second “j” denotes the column number. They say “dimensional matrix”, meaning that the matrix has m rows and n columns. Always in the same matrix
Operations on matrices
Let aij be the elements of matrix A, and bij be the elements of matrix B.
Linear operations:
Multiplying a matrix A by a number λ (symbol: λA) consists of constructing a matrix B, the elements of which are obtained by multiplying each element of the matrix A by this number, that is, each element of the matrix B is equal to
Addition of matrices A + B is the operation of finding a matrix C, all elements of which are equal to the pairwise sum of all corresponding elements of matrices A and B, that is, each element of matrix C is equal to
Subtraction of matrices A − B is defined similarly to addition; this is the operation of finding a matrix C whose elements
Addition and subtraction are only allowed for matrices of the same size.
There is a zero matrix Θ such that adding it to another matrix A does not change A, that is
All elements of the zero matrix are equal to zero.
Nonlinear operations:
Matrix multiplication (designation: AB, less often with a multiplication sign) is the operation of calculating a matrix C, the elements of which are equal to the sum of the products of elements in the corresponding row of the first factor and column of the second.cij = ∑ aikbkj k
The first factor must have the same number of columns as the number of rows in the second. If matrix A has dimension B - , then the dimension of their product AB = C is. Matrix multiplication is not commutative.
Matrix multiplication is associative. Only square matrices can be raised to powers.
Matrix transposition (symbol: AT) is an operation in which the matrix is reflected relative to the main diagonal, that is
If A is a size matrix, then AT is a size matrix
Derivative of a complex function
The complex function has the form: F(x) = f(g(x)), i.e. is a function of a function. For example, y = sin2x, y = ln(x2+2x), etc.
If at the point x the function g(x) has the derivative g"(x), and at the point u = g(x) the function f(u) has the derivative f"(u), then the derivative of the complex function f(g(x)) at point x exists and is equal to f"(u)g"(x).
Implicit function derivative
In many problems, the function y(x) is specified implicitly. For example, for the functions below
it is impossible to obtain the dependence y(x) explicitly.
The algorithm for calculating the derivative y"(x) from an implicit function is as follows:
You first need to differentiate both sides of the equation with respect to x, assuming that y is a differentiable function of x and using the rule for calculating the derivative of a complex function;
Solve the resulting equation for the derivative y"(x).
Let's look at a few examples to illustrate.
Differentiate the function y(x) given by the equation.
Let's differentiate both sides of the equation with respect to the variable x:
what leads to the result
Lapital's rule
L'Hopital's rule. Let the function f(x) and g(x) have in the environment. t-ki x0 pr-nye f' and g' excluding the possibility of this very t-tu x0. Let lim(x®Dx)=lim(x®Dx)g(x)=0 so that f(x)/g(x) for x®x0 gives 0/0. lim(x®x0)f'(x)/g'(x) $ (4), when it coincides with the limit of the ratio of the function lim(x®x0)f(x)/g(x)= lim(x ®x0)f'(x)/g'(x) (5)
44
.1.(Criterion for the monotonicity of a function having a derivative on the interval) Let the function 
continuous on
(a,b), and has a derivative f"(x) at each point. Then
1)f increases by (a,b) if and only if
2) decreases by (a,b) if and only if
2. (Sufficient condition for the strict monotonicity of a function having a derivative on the interval) Let the function 
is continuous on (a,b), and has a derivative f"(x) at each point. Then
1) if then f strictly increases on (a,b);
2) if then f strictly decreases on (a,b).
The converse, generally speaking, is not true. The derivative of a strictly monotonic function can vanish. However, the set of points where the derivative is not zero must be dense on the interval (a,b). More precisely, it does.
3. (Criterion for the strict monotonicity of a function having a derivative on the interval) Let and the derivative f"(x) is defined everywhere on the interval. Then f strictly increases on the interval (a,b) if and only if the following two conditions are satisfied:
Dot product of vectors. Angle between vectors. The condition of parallelism or perpendicularity of vectors.
The scalar product of vectors is the product of their lengths and the cosine of the angle between them:
The following statements are proved in exactly the same way as in planimetry:
The scalar product of two nonzero vectors is zero if and only if the vectors are perpendicular.
The scalar square of a vector, that is, the scalar product of itself and itself, is equal to the square of its length.
The scalar product of two vectors and given by their coordinates can be calculated using the formula
Vectors are perpendicular if and only if their dot product is zero. Example. Given two vectors and . These vectors will be perpendicular if the expression x1x2 + y1y2 = 0. The angle between non-zero vectors is the angle between straight lines for which these vectors are guides. By definition, the angle between any vector and the zero vector is considered equal to zero. If the angle between the vectors is 90°, then such vectors are called perpendicular. We will denote the angle between the vectors as follows:
Instructions
If the original vector is depicted in the drawing in a rectangular two-dimensional coordinate system and a perpendicular one needs to be constructed there, proceed from the definition of perpendicularity of vectors on a plane. It states that the angle between such a pair of directed segments must be equal to 90°. An infinite number of such vectors can be constructed. Therefore, draw a perpendicular to the original vector in any convenient place on the plane, lay a segment on it equal to the length of a given ordered pair of points and assign one of its ends as the beginning of the perpendicular vector. Do this using a protractor and ruler.
If the original vector is given by two-dimensional coordinates ā = (X₁;Y₁), assume that the scalar product of a pair of perpendicular vectors must be equal to zero. This means that you need to select for the desired vector ō = (X₂,Y₂) such coordinates that the equality (ā,ō) = X₁*X₂ + Y₁*Y₂ = 0 will hold. This can be done like this: choose any non-zero value for the X₂ coordinate, and calculate the Y₂ coordinate using the formula Y₂ = -(X₁*X₂)/Y₁. For example, for the vector ā = (15;5) there will be a vector ō, with the abscissa equal to one and the ordinate equal to -(15*1)/5 = -3, i.e. ō = (1;-3).
For a three-dimensional and any other orthogonal coordinate system, the same necessary and sufficient condition for the perpendicularity of vectors is true - their scalar product must be equal to zero. Therefore, if the initial directed segment is given by coordinates ā = (X₁,Y₁,Z₁), select for the ordered pair of points ō = (X₂,Y₂,Z₂) perpendicular to it such coordinates that satisfy the condition (ā,ō) = X₁*X₂ + Y₁*Y₂ + Z₁*Z₂ = 0. The easiest way is to assign single values to X₂ and Y₂, and calculate Z₂ from the simplified equality Z₂ = -1*(X₁*1 + Y₁*1)/Z₁ = -(X₁+Y₁)/ Z₁. For example, for the vector ā = (3,5,4) this will take the following form: (ā,ō) = 3*X₂ + 5*Y₂ + 4*Z₂ = 0. Then take the abscissa and ordinate of the perpendicular vector as one, and in in this case it will be equal to -(3+5)/4 = -2.
Sources:
- find the vector if it is perpendicular
They are called perpendicular vector, the angle between which is 90º. Perpendicular vectors are constructed using drawing tools. If their coordinates are known, then the perpendicularity of the vectors can be checked or found using analytical methods.
You will need
- - protractor;
- - compass;
- - ruler.
Instructions
Construct a vector perpendicular to the given one. To do this, at the point that is the beginning of the vector, restore a perpendicular to it. This can be done using a protractor, setting an angle of 90º. If you don't have a protractor, use a compass to do it.
Set it to the starting point of the vector. Draw a circle with an arbitrary radius. Then construct two with centers at the points where the first circle intersected the line on which the vector lies. The radii of these circles must be equal to each other and larger than the first circle constructed. At the points of intersection of the circles, construct a straight line that will be perpendicular to the original vector at its origin, and plot on it a vector perpendicular to this one.