How to find a vector perpendicular to a given one. Dot product of vectors
ohm. To do this, we first introduce the concept of a segment.
Definition 1
A segment is a part of a straight line that is bounded by points on both sides.
Definition 2
The ends of the segment will be called the points that limit it.
To introduce the definition of a vector, one of the ends of the segment will be called its beginning.
Definition 3
We will call a vector (directed segment) such a segment, for which it is indicated which boundary point is its beginning and which is its end.
Notation: \overline(AB) - vector AB , starting at point A and ending at point B .
Otherwise, in one small letter: \overline(a) (Fig. 1).
Definition 4
Zero vector is any point that belongs to the plane.
Designation: \overline(0) .
We now introduce directly the definition of collinear vectors.
We also introduce the definition of the scalar product, which we will need below.
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 the given 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 the coordinates of the vectors 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 be equal to zero.
Proof.
Need: Let us be given vectors \overline(α) and \overline(β) , which 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 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 has been proven.
Example 1
Prove that the vectors with coordinates (1,-5,2) and (2,1,3/2) are perpendicular.
Proof.
Let's find the dot product for these vectors through 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
Hence, by Theorem 1, these vectors are perpendicular.
Finding a perpendicular vector to two given vectors through the cross product
Let us first introduce the concept of a vector product.
Definition 7
The vector product of two vectors will be called such 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 this vector with two initial ones has the same orientation as the Cartesian coordinate system.
Designation: \overline(α)x\overline(β)x.
To find the vector product, we will use the formula
\overline(α)x\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, then it will be a claim vector. That is, in order to find a vector perpendicular to two vectors, you just need to find their cross product.
Example 2
Find vector perpendicular to vectors with coordinates \overline(α)=(1,2,3) and \overline(β)=(-1,0,3)
Find the cross product of these vectors.
\overline(α)x\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
The condition of perpendicularity of vectors
Vectors are perpendicular if and only if their dot product is zero.
Two vectors a(xa;ya) and b(xb;yb) are given. 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 tasks on a straight line on a plane.
Any straight line on the plane can be given by the first order equation Ax + Vy + C = 0, and the constants A, 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 a straight line. Depending on the values of the constants A, B and C, the following special cases are possible:
C \u003d 0) - the straight line is parallel to the Ox axis - B \u003d 0, A 0, C 0 (Ax + C \u003d 0) - the straight line is parallel to the Oy axis - B \u003d C \u003d 0, A 0 - the 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 ur-th Ax+By+C=0 is equal to 0, ur-e
called incomplete. By the form of the equation of a straight line, one can judge its position on
damn ohh. Possible cases:
1 C=0 L: Ax+By=0 t. O(0,0) satisfies this equation, which means the line
passes through the origin
2 А=0 L: Ву+С=0 - normal v-p n=(0,B) is perpendicular to the OX axis from here
it follows that the line is parallel to the x-axis
3 V \u003d 0 L: Ay + C \u003d 0 0 - normal v-r n \u003d (A, 0) is perpendicular to the OY axis from here
it follows that the line is parallel to the y-axis
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 properties are:
1. If you rearrange two rows (or two columns) of the determinant, then 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 the rows and columns are swapped, preserving their order.
4. If all elements of any 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.
Matrix and action on them
Matrix- a mathematical object written as a rectangular table of numbers (or ring elements) and allowing algebraic operations (addition, subtraction, multiplication, etc.) between it and other similar objects. Usually matrices are represented by two-dimensional (rectangular) tables. Sometimes multidimensional matrices or non-rectangular matrices are considered.
Usually, the matrix is denoted by a capital letter of the Latin alphabet and is distinguished by round brackets “(…)” (there is also a selection of 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 element of the matrix has 2 subscripts (aij) - the first "i" indicates the number of the row in which the element is located, and the second "j" is the number of the column. They say "dimension matrix", meaning that the matrix has m rows and n columns. Always in the same matrix
Matrix operations
Let aij be elements of matrix A and bij be elements of matrix B.
Linear operations:
Multiplication of a matrix A by a number λ (notation: λA) consists in 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
The addition of matrices A + B is the operation of finding a matrix C, all of whose elements are equal to the pairwise sum of all the 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, it is the operation of finding a matrix C whose elements
Addition and subtraction are allowed only for matrices of the same size.
There is a zero matrix Θ such that its addition to another matrix A does not change A, i.e.
All elements of the zero matrix are equal to zero.
Nonlinear operations:
Matrix multiplication (notation: AB, rarely with a multiplication sign) is an operation to calculate a matrix C, the elements of which are equal to the sum of the products of the elements in the corresponding row of the first factor and the column of the second.cij = ∑ aikbkj k
The first multiplier must have as many columns as there are rows in the second. If the 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 a power.
Matrix transposition (symbol: AT) is an operation in which the matrix is reflected along the main diagonal, i.e.
If A is a size matrix, then AT is a size matrix
Derivative of a compound 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) is the derivative g "(x), and at the point u \u003d g (x) the function f (u) has the derivative f" (u), then the derivative of the complex function f (g (x)) in point x exists and is equal to f"(u)g"(x).
Derivative of an implicit function
In many problems, the function y(x) is specified in an indirect way. For example, for the functions below
it is impossible to obtain the dependence y(x) explicitly.
The algorithm for calculating the derivative y "(x) of an implicit function is as follows:
First, you 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 with respect to the derivative y "(x).
Let's look at a few examples to illustrate.
Differentiate the function y(x) given by the equation.
Differentiate both sides of the equation with respect to the variable x:
which leads to the result
Lapital's rule
L'Hopital's rule. Let f-tion f(x) and g(x) have in the env. t-ki x0 pr-nye f‘ and g‘ excluding the possibility of this very t-ku 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.(A criterion for the monotonicity of a function that has a derivative on an interval) Let the function 
continuous on
(a,b), and has a derivative f"(x) at every point. Then
1)f increases by (a,b) if and only if
2) decreases on (a,b) if and only if
2. (A sufficient condition for the strict monotonicity of a function that has a derivative on an interval) Let the function 
is continuous on (a,b), and has a derivative f"(x) at every point. Then
1) if then f is strictly increasing on (a,b);
2) if then f is strictly decreasing on (a,b).
The converse is generally not true. The derivative of a strictly monotonic function can vanish. However, the set of points where the derivative is not equal to zero must be dense on the interval (a,b). More precisely, it takes place.
3. (A criterion for the strict monotonicity of a function that has a derivative on an 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:
Scalar product of vectors. Angle between vectors. 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:
In exactly the same way as in planimetry, the following assertions are proved:
The scalar product of two non-zero vectors is zero if and only if these vectors are perpendicular.
The dot square of a vector, i.e. the dot 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 by 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 the lines for which these vectors are guides. The angle between any vector and a zero vector is, by definition, considered equal to zero. If the angle between vectors is 90°, then such vectors are called perpendicular. The angle between vectors will be denoted as follows:
Instruction
If the original vector is shown in the drawing in a rectangular two-dimensional coordinate system and a perpendicular one needs to be built in the same place, 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°. It is possible to construct an infinite number of such vectors. Therefore, draw a perpendicular to the original vector in any convenient place on the plane, set aside on it a segment 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 with a protractor and a ruler.
If the original vector is given by two-dimensional coordinates ā = (X₁;Y₁), proceed from the fact that the scalar product of a pair of perpendicular vectors must be equal to zero. This means that you need to choose for the desired vector ō = (X₂,Y₂) such coordinates at which 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 original directed segment is given by the coordinates ā = (X₁,Y₁,Z₁), for the ordered pair of points ō = (X₂,Y₂,Z₂) perpendicular to it, choose 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 equation 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 unity, and in in this case will be equal to -(3+5)/4 = -2.
Sources:
- find vector if it is perpendicular
Perpendicular are called vector, the angle between which is 90º. Perpendicular vectors are built using drawing tools. If their coordinates are known, then it is possible to check or find the perpendicularity of the vectors by analytical methods.
You will need
- - protractor;
- - compass;
- - ruler.
Instruction
Set it to the start point of the vector. Draw a circle with an arbitrary radius. Then build two centered at the points where the first circle intersects the line that the vector lies on. The radii of these circles must be equal to each other and greater than the first constructed circle. At the intersection points of the circles, construct a straight line that will be perpendicular to the original vector at the point of its beginning, and set aside on it a vector perpendicular to the given one.
Find a vector perpendicular to the one whose coordinates and are equal to (x; y). To do this, find a pair of numbers (x1;y1) that would satisfy the equality x x1+y y1=0. In this case, the vector with coordinates (x1;y1) will be perpendicular to the vector with coordinates (x;y).
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 related to the equations of lines and planes.
We will consider the necessary and sufficient condition for two vectors to be perpendicular, decide on the method of finding a vector perpendicular to a given one, and touch on situations in finding a vector that is perpendicular to two vectors.
Yandex.RTB R-A-339285-1
Necessary and sufficient condition for two vectors to be perpendicular
Let's apply the rule about perpendicular vectors on the plane and in three-dimensional space.
Definition 1
Given the value of the angle between two non-zero vectors 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?
The establishment of perpendicularity is possible through the drawing. When plotting a vector on a plane from given points, you can geometrically measure the angle between them. The perpendicularity of the vectors, if it is established, is not entirely accurate. Most often, these problems do not allow you to do this with a protractor, so this method is only applicable when nothing else is known about vectors.
Most cases of proving the perpendicularity of two non-zero vectors on a plane or in space is done using necessary and sufficient condition for perpendicularity of two vectors.
Theorem 1
The scalar product of two non-zero vectors a → and b → equal to zero to fulfill the equality a → , b → = 0 is sufficient for their perpendicularity.
Proof 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, and, therefore, 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 .
The second part of the proof
Under the condition when a ⇀ , b → = 0 prove the perpendicularity of a → and b → .
In fact, the proof is the reverse of the previous one. It is known that a → and b → are non-zero, so 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 90 ° . By definition, this is a necessary and sufficient property.
Perpendicular condition on the coordinate plane
Chapter dot 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. A necessary and sufficient condition for two vectors to be perpendicular 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 by condition it will be 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 satisfied, which means that the given vectors are perpendicular on the plane.
Answer: yes, the given vectors a → and b → are perpendicular.
Example 2
Given coordinate vectors i → , j → , k → . Check if vectors i → - j → and i → + 2 j → + 2 k → can be perpendicular.
Solution
In order to remember how the coordinates of a vector are determined, you need to read an article about vector coordinates in rectangular coordinates. Thus, we obtain that the given vectors i → - j → and i → + 2 j → + 2 k → have the corresponding coordinates (1, - 1, 0) and (1, 2, 2) . Substitute the numerical values and get: i → + 2 j → + 2 k → , i → - j → = 1 1 + (- 1) 2 + 0 2 = - 1 .
The expression is not 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 because the condition is not satisfied.
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 λ for which the given 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. With 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 \u003d 8, A C \u003d 6, B C \u003d 10 cm. Check the vectors A B → and A C → for perpendicularity.
Solution
When the vectors A B → and A C → are perpendicular, the triangle A B C is considered rectangular. Then we apply the Pythagorean theorem, where BC is the hypotenuse of the triangle. The equality B C 2 = A B 2 + A C 2 must be satisfied. It follows that 10 2 = 8 2 + 6 2 ⇔ 100 = 100 . Hence, A B and A C are the 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 in the plane. Let's represent it on the coordinate line.
A non-zero vector a → , lying on the line a, is given. Then the given b → , located on any line perpendicular to the line a, becomes perpendicular and 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 except zero, then finding the coordinates of the vector b → perpendicular to a → = (a x , a y) reduces 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 nonzero and b x is computed from the inequality a x · b x + a y · b y = 0 ⇔ b x = - a y · b y a x . When a x = 0 and a y ≠ 0, we assign b x any value other than zero, and b y is found 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 the given one.
Solution
Denote the desired vector as b → (b x , b y) . You can find its coordinates 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 . 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 . Hence, the vector b → = (1 2 , 1) is a vector perpendicular to a → .
Answer: b → = (1 2 , 1) .
If the question of three-dimensional space is raised, the problem is solved according to the same principle. For a given vector a → = (a x , a y , a z) there is an infinite set of perpendicular vectors. Will fix it on the coordinate 3D plane. Given a → lying on the line a . The plane perpendicular to the straight line 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 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 hold. From the condition a → - non-zero, which means that one of the coordinates has a value not equal to zero. Suppose 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 get 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 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 with an example.
Example 6
Given a vector with coordinates a → = (1 , 2 , 3) . Find a vector perpendicular to the given one.
Solution
Denote the desired vector as 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 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) . The vector b → is one of the perpendicular vectors to the given one.
Answer: b → = (- 5 , 1 , 1) .
Finding the coordinates of a vector perpendicular to two given vectors
You 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) . Under the condition that the vectors a → and b → are collinear, in the problem it will be enough to find a vector perpendicular to a → or b → .
When solving, the concept of a vector product of vectors is used.
Cross 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 us analyze the vector product in more detail using the example of the problem.
Example 7
Vectors b → = (0 , 2 , 3) and a → = (2 , 1 , 0) are given. Find the coordinates of any perpendicular vector to the data at the same time.
Solution
To solve, you need to find the cross product of vectors. (Must refer to paragraph matrix determinant calculations 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 given a → and b → .
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