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 aside 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.
The unit vector is: , where 
– vector module.
Answer: 
.
Note. The coordinates of the unit vector must be no more than one.
6.3. Find the length and direction cosines of a vector 
. Compare with the answer in the previous paragraph. Draw conclusions.
The length of a vector is its modulus:
And we can find the direction cosines using the formula for one of the ways to specify vectors:
From this we see that the direction cosines are the coordinates of the unit vector.
Answer: 
,
,
,
.
6.4. Find 
.
It is necessary to perform the actions of multiplying a vector by a number, adding and modulus.
We multiply the coordinates of the vectors by a number term by term.
We add the coordinates of the vectors term by term.
Finding the modulus of the vector.
Answer: 
6.5. Determine vector coordinates 
, collinear to the vector 
, knowing that 
and it is directed in the direction opposite to the vector 
.
Vector 
collinear to the vector 
, which means its unit vector is equal to the unit vector 
only with a minus sign, because directed in the opposite direction.
The unit vector has a length equal to 1, which means that if you multiply it by 5, then its length will be equal to five.
We find 
Answer: 
6.6. Calculate Dot Products 
And 
. Are the vectors perpendicular? 
And 
,
And 
between themselves?
Let's do the scalar product of vectors.
If the vectors are perpendicular, their scalar product is zero.
We see that in our case the vectors 
And 
perpendicular.
Answer: 
,
, the vectors are not perpendicular.
Note. The geometric meaning of the scalar product is of little use in practice, but it still exists. The result of such an action can be depicted and calculated geometrically.
6.7. Find the work done by a material point to which a force is applied 
, when moving it from point B to point C.
The physical meaning of the scalar product is work. The force vector is here 
, the displacement vector is 
. And the product of these vectors will be the required work.
Finding a job
6.8. Find the interior angle at a vertex A and external vertex angle C triangle ABC .
From the definition of the scalar product of vectors, we obtain the formula for finding the angle: .
IN 
We will look for the interior angle as the angle between vectors emanating from one point.
To find the external angle, you need to combine the vectors so that they come out from one point. The picture explains this.
It is worth noting that 
, just have different initial coordinates.
Finding the necessary vectors and angles
Answer: internal angle at vertex A = 
, external angle at vertex B = 
.
6.9. Find the projections of the vectors: and
Let us recall the vector vectors: 
,
,
.
The projection is also found from the scalar product
-projection b on a.
Previously received vectors
,
,
Finding the projection
Finding the second projection
Answer: 
,
Note. The minus sign when finding a projection means that the projection does not descend onto the vector itself, but in the opposite direction, onto the line on which this vector lies.
6.10. Calculate 
.
Let's do the vector product of vectors
Let's find the module
We find the sine of the angle between vectors from the definition of the vector product of vectors
Answer: 
,
,
.
6.11. Find the area of a triangle ABC and the length of the height descended from point C.
The geometric meaning of the modulus of a vector product is that it is the area of the parallelogram formed by these vectors. And the area of a triangle is equal to half the area of a parallelogram.
The area of a triangle can also be found as the product of the height and the base divided by two, from which the formula for finding the height can be derived.
Thus, we find the height
Answer: 
,
.
6.12. Find the unit vector perpendicular to the vectors 
And 
.
The result of the dot product is a vector that is perpendicular to the two original ones. And a unit vector is a vector divided by its length.
Previously, we found:
,
Answer: 
.
6.13. Determine the magnitude and direction cosines of the moment of force 
, applied to A relative to point C.
The physical meaning of a vector product is the moment of force. Let's give an illustration for this task.
Finding the moment of force
Answer: 
.
6.14. Do the vectors lie 
,
And 
in the same plane? Can these vectors form the basis of space? Why? If they can, expand the vector into this basis 
.
To check whether the vectors lie in the same plane, it is necessary to perform a mixed product of these vectors.
The mixed product is not equal to zero, therefore, the vectors do not lie in the same plane (not coplanar) and can form a basis. Let's decompose 
on this basis.
Let us expand by basis by solving the equation
Answer: Vectors 
,
And 
do not lie in the same plane. 
.
6.15. Find 
. What is the volume of the pyramid with vertices A, B, C, D and its height lowered from point A to the base BCD.
G 
The geometric meaning of the mixed product is that it is the volume of the parallelepiped formed by these vectors.
The volume of the pyramid is six times less than the volume of the parallelepiped.
The volume of the pyramid can also be found like this:
We get the formula for finding the height
Finding the height
Answer: volume = 2.5, height = 
.
6.16. Calculate 
And 
.
– We invite you to think about this task yourself.
- Let's perform the work.
Previously received
Answer: 
.
6.17. Calculate
Let's do the steps in parts
3)
Let's sum up the obtained values
Answer: 
.
6.18. Find vector 
, knowing that it is perpendicular to the vectors 
And 
, and its projection onto the vector 
equals 5.
Let's split this task into two subtasks
1) Find a vector perpendicular to the vectors 
And 
arbitrary length.
We get the perpendicular vector as a result of the vector product
Previously, we found:
The required vector differs only in length from the received one
2) Let's find 
through the equation
6.19. Find vector 
, satisfying the conditions 
,
,
.
Let us consider these conditions in more detail.
This is a system of linear equations. Let's compose and solve this system.
Answer: 
6.20. Determine the coordinates of a vector 
, coplanar with the vectors 
And 
, and perpendicular to the vector 
.
In this task there are two conditions: coplanarity of vectors and perpendicularity; let’s first fulfill the first condition, and then the second.
1) If the vectors are coplanar, then their mixed product is equal to zero.
From here we obtain some dependence of the coordinates of the vector
Let's find the vector 
.
2) If the vectors are perpendicular, then their scalar product is zero
We have obtained the second dependence of the coordinates of the desired vector
For any value 
the vector will satisfy the conditions. Let's substitute 
.
Answer: 
.
Analytic geometry
In the section on the question, find a vector perpendicular to two given vectors given by the author Anna Afanasyeva the best answer is: A vector perpendicular to two non-parallel vectors is found as their vector product xb, to find it you need to compose a determinant, the first line of which will consist of the unit vectors I, j, k, the second from the coordinates of vector a, the third from the coordinates of vector b . The determinant is considered to be an expansion along the first line, in your case you get akhv=20i-10k, or ahv=(20,0,-10).
Answer from 22 answers[guru]
Hello! Here is a selection of topics with answers to your question: find a vector perpendicular to two given vectors
Answer from stretch out[newbie]
A vector perpendicular to two non-parallel vectors is found as their vector product xb, to find it you need to compose a determinant, the first line of which will consist of the unit vectors I, j, k, the second - from the coordinates of vector a, the third - from the coordinates of vector b. The determinant is considered to be an expansion along the first line, in your case you get akhv=20i-10k, or ahv=(20,0,-10).
Answer from HIKA[guru]
Roughly decide this way; But first, read everything yourself!! !
Calculate the scalar product of vectors d and r if d=-c+a+2b; r=-b+2a.
The modulus of vector a is 4, the modulus of vector b is 6. The angle between vectors a and b is 60 degrees, vector c is perpendicular to vectors a and b.
Points E and F lie respectively on sides AD and BC of parallelogram ABCD, with AE = ED, BF: FC = 4: 3. a) Express the vector EF in terms of vectors m = vector AB and vector n = vector AD. b) Can the equality vector EF = x multiplied by the vector CD hold for any value of x? .
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.
Yandex.RTB R-A-339285-1
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 a 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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ohm To do this, we first introduce the concept of a segment.
Definition 1
We will call a segment a part of a line that is bounded by points on both sides.
Definition 2
The ends of a segment are the points that limit it.
To introduce the definition of a vector, we call one of the ends of the segment its beginning.
Definition 3
We will call a vector (directed segment) a segment in which it is indicated which boundary point is its beginning and which is its end.
Notation: \overline(AB) is a vector AB that starts at point A and ends at point B.
Otherwise, in one small letter: \overline(a) (Fig. 1).
Definition 4
We will call the zero vector any point that belongs to the plane.
Notation: \overline(0) .
Let us now introduce directly the definition of collinear vectors.
We will also introduce the definition of a scalar product, which we will need later.
Definition 6
The scalar product of two given vectors is a scalar (or number) that is equal to the product of the lengths of these two vectors with the cosine of the angle between these vectors.
Mathematically it might look like this:
\overline(α)\overline(β)=|\overline(α)||\overline(β)|cos∠(\overline(α),\overline(β))
The dot product can also be found using vector coordinates as follows
\overline(α)\overline(β)=α_1 β_1+α_2 β_2+α_3 β_3
Sign of perpendicularity through proportionality
Theorem 1
For non-zero vectors to be perpendicular to each other, it is necessary and sufficient that their scalar product of these vectors equals zero.
Proof.
Necessity: Let us be given vectors \overline(α) and \overline(β) that have coordinates (α_1,α_2,α_3) and (β_1,β_2,β_3), respectively, and they are perpendicular to each other. Then we need to prove the following equality
Since the vectors \overline(α) and \overline(β) are perpendicular, the angle between them is 90^0. Let's find the scalar product of these vectors using the formula from Definition 6.
\overline(α)\cdot \overline(β)=|\overline(α)||\overline(β)|cos90^\circ =|\overline(α)||\overline(β)|\cdot 0=0
Sufficiency: Let the equality be true \overline(α)\cdot \overline(β)=0. Let us prove that the vectors \overline(α) and \overline(β) will be perpendicular to each other.
By definition 6, the equality will be true
|\overline(α)||\overline(β)|cos∠(\overline(α),\overline(β))=0
Cos∠(\overline(α),\overline(β))=0
∠(\overline(α),\overline(β))=90^\circ
Therefore, the vectors \overline(α) and \overline(β) will be perpendicular to each other.
The theorem is proven.
Example 1
Prove that vectors with coordinates (1,-5,2) and (2,1,3/2) are perpendicular.
Proof.
Let's find the scalar product for these vectors using the formula given above
\overline(α)\cdot \overline(β)=1\cdot 2+(-5)\cdot 1+2\cdot \frac(3)(2)=2\cdot 5+3=0
This means, according to Theorem 1, these vectors are perpendicular.
Finding a perpendicular vector to two given vectors using the cross product
Let us first introduce the concept of a vector product.
Definition 7
The vector product of two vectors will be a vector that will be perpendicular to both given vectors, and its length will be equal to the product of the lengths of these vectors with the sine of the angle between these vectors, and also this vector with two initial ones has the same orientation as the Cartesian coordinate system.
Designation: \overline(α)х\overline(β) x.
To find the vector product, we will use the formula
\overline(α)х\overline(β)=\begin(vmatrix)\overline(i)&\overline(j)&\overline(k)\\α_1&α_2&α_3\\β_1&β_2&β_3\end(vmatrix) x
Since the vector of the cross product of two vectors is perpendicular to both of these vectors, it will be the vector. That is, in order to find a vector perpendicular to two vectors, you just need to find their vector product.
Example 2
Find a vector perpendicular to vectors with coordinates \overline(α)=(1,2,3) and \overline(β)=(-1,0,3)
Let's find the vector product of these vectors.
\overline(α)х\overline(β)=\begin(vmatrix)\overline(i)&\overline(j)&\overline(k)\\1&2&3\\-1&0&3\end(vmatrix)=(6- 0)\overline(i)-(3+3)\overline(j)+(0+2)\overline(k)=6\overline(i)-6\overline(j)+2\overline(k) =(6,6,2) x