No. 1. Let's find the vectors a and b: a = α m + β n = -5 m - 7 n b = γ m + δ n = -3 m + 2 n The values of the constants are also given: k = 2, ℓ = 11, φ = 3π/2, α = -5, β = -7, γ = -3, δ = 2, λ = -3, μ = 4, ν = -1, τ = 2.
а) Найдем ( λ·a + μ·b );( ν·a + τ·b ): λ·a + μ·b = -3(-5m - 7n) + 4(-3m + 2n) = 15m - 17n ν·a + τ·b = -1(-5m - 7n) + 2(-3m + 2n) = 11m - 2n ( λ·a + μ·b )·( ν·a + τ·b ) = ( 15m - 17n); (11m - 2n) = -352
b) Find the projection ( ν a + τ b ) onto b: ( ν a + τ b ) = (-1)(-5m - 7n) + 2(-3m + 2n) = -11m + 4n Projection vector to another vector is equal to the scalar product of the vector and the unit vector of the direction of this vector, i.e.: proj_b(ν·a + τ·b) = ((-11m + 4n)·(-3m + 2n))/|b| = (-23/13)(-3m + 2n)
в) Single cos( a + τ·b ): a + τ·b = (-5m - 7n) + 2(-3m + 2n) = -11m - 3n cos( a + τ·b ) = cos(arccos( (a+b)/|a+b|)) = cos(arccos((-11m - 3n)/sqrt(170))) = (-11sqrt(170))/170 - (3sqrt(170))/170 = -14sqrt(170)/1
No. 2. Let's find the vectors a, b, c and d: a = AB = B - A = (-8; -6; 3) b = AC = C - A = (2; 1; -5) c = [a, b] = a x b = (-27; 28; 2) d = AM = A + α·(B - A) = (4; 3; 2) - α(8; 6; -3)
The values of the constants are also given: A(4;3;2), B(-4;-3;5), C(6;4;-3), α.
a) Modulus of vector a: |a| = sqrt((-8)^2 + (-6)^2 + 3^2) = sqrt(109)
b) Scalar product of vectors a and b: a·b = (-8·2) + (-6·1) + (3·(-5)) = -29
c) Projection of vector c onto vector d: proj_d(c) = (c·d)/|d|^2 = ((-27·(-4α)) + (28·6α) + (2·(-3α) ))/((8α)^2 + (6α)^2 + (-3α)^2) = (-8α)/13
d) Coordinates of the point M dividing the segment ℓ in relation to α: AM/AB = α/1 M = A + α·AB = (4; 3; 2) + α(-8; -6; 3) = (-4α + 4; -3α + 3; 3α + 2)
ℓ is not given, so the answer depends on the specific value of ℓ.
No. 3. To prove that the vectors a, b, and c form a basis, one must show that they are linearly independent and that any vector in space can be expressed as a linear combination of these vectors.
Let's check the linear independence of vectors a, b and c. To do this, we create the equation a x + b y + c z = 0 and show that its only solution is x = y = z = 0.
a·x + b·y + c·z = 0 (-2x + 3y + 4z; 5x + 2y - 3z; x - y + 2z) + (3u - 2v; 2u - 3v; -u + v) + (4p; -q; 2p) = 0 (-2x + 3y + 4z + 3u - 2v + 4p; 5x + 2y - 3z + 2u - 3v - q; x - y + 2z - u + v + 2p) = 0
This is a system of three equations with three unknowns. Applying the Gaussian method, we obtain a unique solution x = y = z = u = v = p = q = 0. This means that the vectors a, b and c are linearly independent, that is, they form a basis in space.
Let us now find the coordinates of the vector d in this basis. To do this, we express vector d in terms of vectors a, b and c:
d = α·a + β·b + γ·c
Let's substitute the known values into the formula:
d = α(-2; 5; 1) + β(3; 2; -1) + γ(4; -3; 2) = (-2α + 3β + 4γ; 5α + 2β - 3γ; α - β + 2c)
Now we need to solve the system of equations:
-2a + 3b + 4c = -4 5a + 2b - 3c = 22 a - b + 2c = -13
Applying the Gaussian method, we obtain the solution: α = 1, β = -4, γ = 3. This means that the coordinates of the vector d in the basis {a, b, c} are equal to (1; -4; 3).
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No. 1. In this problem you need to find: a) the value of the expression ( λ·a + μ·b )·( ν·a + τ·b ), which is the scalar product of two linear combinations of vectors a and b; b) projection of the vector ( ν·a + τ·b ) onto the vector b; c) the value of cos(a + τ·b), where a and b are given vectors.
To solve the problem, you need to use the formulas for the scalar product, the projection of a vector onto another vector, and the vector addition formula. As initial data, the coefficients α, β, γ, δ, k, ℓ, φ, λ, μ, ν and τ are given, which must be used in the calculations.
No. 2. In this problem you need to find: a) modulus of vector a; b) scalar product of vectors a and b; c) projection of vector c onto vector d; d) coordinates of the point M that divides the segment ℓ in a given ratio α.
To solve the problem, you need to use formulas for the modulus of a vector, scalar product, projection of a vector onto another vector, as well as formulas for finding the coordinates of a point dividing a segment in a given ratio. The initial data are the coordinates of points A, B and C, as well as the necessary coefficients for calculations.
No. 3. In this problem, you need to prove that vectors a, b, c form a basis, and find the coordinates of vector d in this basis.
To solve the problem, you need to show that the vectors a, b, c are linearly independent and that any vector in three-dimensional space can be represented as a linear combination of these vectors. Next, you need to find the coordinates of vector d in the basis a, b, c, using formulas for finding the coefficients of a linear combination. The initial data are vectors a, b, c and vector d, which must be used when solving the problem.
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Solution to problem 21.1.11 from the collection of Kepe O.E. - 21.1.11 Однородный цилиндр массой 2 кг может катиться по... ...