IDZ Ryabushko 2.2 Option 6

Vector data:
- a(3;-2;1);
- b(0;2;-3);
- c(-3;2;-1).
Necessary:
- a) calculate the mixed product of three vectors;
- b) find the modulus of the vector product;
- c) calculate the scalar product of two vectors;
- d) check whether two vectors are collinear or orthogonal;
- e) check whether the three vectors are coplanar.
Answer:
- a) The mixed product of vectors a, b and c is calculated by the formula:
(a × b) ⋅ c = (b × c) ⋅ a = (c × a) ⋅ b = a₁(b₂c₃ − b₃c₂) + a₂(b₃c₁ − b₁c₃) + a₃(b₁c₂ − b₂c₁) = 3(2×(-1) - 2×(-3)) - 2(0×(-1) - (-3)×(-3)) + 1(0×2 - 2×(-3)) = -12.
b) The modulus of the vector product of vectors a and b is equal to:

|a × b| = √(a₂b₃ - a₃b₂)² + (a₃b₁ - a₁b₃)² + (a₁b₂ - a₂b₁)² = √((-2)² + 3² + 6²) = √49 = 7.

c) The scalar product of vectors a and b is calculated by the formula:

a ⋅ b = a₁b₁ + a₂b₂ + a₃b₃ = 3×0 + (-2)×2 + 1×(-3) = -7.

d) Two non-zero vectors will be collinear if one is a multiple of the other. Two non-zero vectors will be orthogonal if their dot product is zero. Let's check:

vectors a and b are not multiples, since their absolute values are not equal and their scalar product is not zero;
vectors a and c are not multiples, since their absolute values are not equal and their scalar product is not zero;
vectors b and c are not multiples, since their absolute values are not equal and their scalar product is not zero.

Therefore, neither two of the three vectors are collinear, nor two vectors are orthogonal.

e) Three vectors will be coplanar if their mixed product is zero. Let's check:

a ⋅ (b × c) = 3×(2×(-1) - 2×(-3)) + (-2)×(0×(-1) - (-3)×(-3)) + 1×(0×2 - 2×(-3)) = -12 ≠ 0.

Therefore, the three vectors are not coplanar.

The tops of the pyramid are located at the points:

A(3;4;2);
B(–2;3;–5);
C(4;–3;6);
D(6;–5;3).

Answer:

To solve the problem you need to find the height of the pyramid and the area of the base.

Let's find the vectors AB, AC and AD:

AB = B - A = (-2 - 3; 3 - 4; -5 - 2) = (-5; -1; -7);
AC = C - A = (4 - 3; -3 - 4; 6 - 2) = (1; -7; 4);
AD = D - A = (6 - 3; -5 - 4; 3 - 2) = (3; -9; 1).

The height of the pyramid lowered to the base ABCD is equal to the length of the projection of the vector AD onto the straight line passing through points B and C. Let us find it:

Let's find the vector product of vectors AB and AC:

AB × AC = (-1×4 - (-7)×1; (-7)×1 - (-5)×4; (-5)×(-1) - (-1)×(-7)) = (-11; -29; -34).

Let's find the vector product of the vectors AB × AC and AC:

(AB × AC) × AC = (-29×4 - (-34)×(-7); (-34)×1 - (-11)×4; (-11)×(-7) - (- 29)×1) = (19; 110; 208).

Let's find the projection of the vector AD onto the vector AB × AC:

proj_AB×ACAD = (AD ⋅ (AB × AC)) / |AB × AC| = (3×19 - 9×110 + 208) / √(19² + 110² + 208²) ≈ 7.585.

Now let's find the area of the base ABCD. To do this, we find the modulus of the vector product of the vectors AB and AC:

|AB × AC| = √((-1)² + (-7)² + 4²) ≈ 7.681.

The area of the base is:

S_grounds = |AB × AC| / 2 ≈ 3.840.

Thus, the height of the pyramid is approximately 7.585, and the area of the base is approximately 3.840.

Force F(3;–5;7) is applied to point A(2;3;–5). Necessary:

a) calculate the work of force in the case when the point of its application, moving rectilinearly, moves to point B(0;4;3);
b) find the modulus of the moment of force relative to point B.

Answer:

a) The work done by force F when moving point A to point B is calculated by the formula:

W = F ⋅ AB = (F, AB) = F₁AB₁ + F₂AB₂ + F₃AB₃ = 3×(-2) + (-5)×1 + 7×8 = 49.

Therefore, the work done by force F is 49.

b) The moment of force F relative to point B is equal to the vector product

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IDZ Ryabushko 2.2 Option 6 is a task in linear algebra, which consists of three tasks:

Given vectors a(3;-2;1), b(0;2;-3) and c(-3;2;-1). You need to do the following:

a) Calculate the mixed product of three vectors. b) Find the modulus of the vector product. c) Calculate the scalar product of two vectors. d) Check whether two vectors are collinear or orthogonal. e) Check whether the three vectors are coplanar.
The vertices of the pyramid are defined by points A(3;4;2), B(-2;3;-5), C(4;-3;6) and D(6;-5;3). It is necessary to find the volume of this pyramid.
Force F(3;-5;7) is applied to point A(2;3;-5). You need to do the following:

a) Calculate the work of the force in the case when the point of its application, moving rectilinearly, moves to point B(0;4;3). b) Calculate the modulus of the moment of force relative to point B.

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