Option 12 IDZ 2.2

No. 1.12. For a given set of vectors, you must perform the following steps:

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 three vectors are coplanar: a(-4;3;-7);b(4;6;-2);c(6;9;-3).

Solution: a) To calculate the mixed product of three vectors, it is necessary to find the determinant of the matrix composed of the coordinates of these vectors: $$\begin{vmatrix} -4 & 3 & -7 \\ 4 & 6 & -2 \\ 6 & 9 & - 3 \end{vmatrix} = (-4) \cdot 6 \cdot (-3) + 3 \cdot (-2) \cdot 6 + (-7) \cdot 4 \cdot 9 - (-7) \cdot 6 \cdot (-2) - 4 \cdot (-2) \cdot (-3) - 3 \cdot 4 \cdot 9 = -84.$$ Answer: -84. b) To find the modulus of the vector product of vectors a and b, it is necessary to calculate the length of the vector obtained as a result of their vector product. The formula for calculating the modulus of a vector product is: $$|a \times b| = |(-39;14;30)| = \sqrt{(-39)^2 + 14^2 + 30^2} \approx 42.01.$$ Answer: 42.01. c) To calculate the scalar product of two vectors, it is necessary to multiply the corresponding coordinates of these vectors and add the resulting products: $$a \cdot b = (-4) \cdot 4 + 3 \cdot 6 + (-7) \cdot (-2) = 8.$$ Answer: 8. d) Two non-zero vectors will be collinear if one of them is a multiple of the other. Two non-zero vectors will be orthogonal if their dot product is zero. Let's calculate the scalar product of vectors a and b: $$a \cdot b = 8 \neq 0,$$ means that vectors a and b are not orthogonal. Next, let's find the vector product of vectors a and b: $$a \times b = (-39;14;30).$$ Let's calculate the scalar product of vectors a and c: $$a \cdot c = (-4) \cdot 6 + 3 \cdot 9 + (-7) \cdot (-3) = 51,$$ means that vectors a and c are not orthogonal. Next, let's find the vector product of vectors a and c: $$a \times c = (-12;-34;-6).$$ Let's calculate the scalar product of vectors b and c: $$b \cdot c = 4 \cdot 6 + 6 \cdot 9 + (-2) \cdot (-3) = 72,$$ means that vectors b and c are not orthogonal. Next, we find the vector product of vectors b and c: $$b \times c =(33;26;18).$$ Thus, neither two nor three vectors are orthogonal, and therefore cannot be coplanar. e) Answer: None of the three vectors is coplanar with the other two vectors. No. 2.12. The vertices of the pyramid are given: A(7;4;9), B(1;-2;-3), C(-5;-3;0), D(1;-3;4). Solution: To find the volume of the pyramid formed by vertices A, B, C and D, it is necessary to find half the volume of the parallelepiped formed by the vectors AB, AC and AD. The volume of a parallelepiped can be calculated using the formula: $$V = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|/6.$$ Calculations: $$\vec{AB} = ( -6;-6;-12),$$ $$\vec{AC} = (-12;-7;-9),$$ $$\vec{AD} = (-6;-7;-5 ),$$ $$\vec{AC} \times \vec{AD} = (-26;78;-42),$$ $$\vec{AB} \cdot (\vec{AC} \times \vec {AD}) = 936.$$ $$V = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|/6 = 156.$$ Answer: volume of the pyramid formed by the vertices A, B, C and D is equal to 156. No. 3.12. Given is the force F(2;2;9) applied to point A(4;2;-3). Solution: a) To calculate the work of force F, it is necessary to find the scalar product of the force vector and the displacement vector of the point of application of the force: $$W = \vec{F} \cdot \vec{s},$$ where $\vec{F}$ - force vector, $\vec{s}$ is the displacement vector of the force application point. The displacement vector of the force application point: $$\vec{s} = \vec{AB} = (-2;2;3),$$ where point B(2;4;0). Then the work of force F when moving the point of application of force from point A to point B is equal to: $$W = \vec{F} \cdot \vec{s} = 14.$$ Answer: work of force F when moving the point of application of force from point A to point B is equal to 14. b) Moment of force F relative B can be calculated using the formula: $$\vec{M_B} = \vec{r_{AB}} \times \vec{F},$$ where $\vec{r_{AB}}$ is the radius vector of the point of application of the force relative to point B. The radius vector of the point of application of force relative to point B: $$\vec{r_{AB}} = \vec{BA} = (2;-2;-3).$$ Then the moment of force F relative to the point B is equal to: $$\vec{M_B} = \vec{r_{AB}} \times \vec{F} = (-22;22;8).$$ Moment modulus is: $$|\vec{M_B} | = \sqrt{(-22)^2 + 22^2 + 8^2} \approx 30.33.$$ Answer: the modulus of the moment of force F relative to point B is approximately approximately

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By purchasing this product, you receive useful and high-quality material that will help you successfully master the topic of linear and vector algebra, as well as prepare for an exam or test.

This product is a collection of solutions to problems in linear and vector algebra, including finding the mixed, scalar and vector product of vectors, determining the collinearity and orthogonality of vectors, calculating the volume of the pyramid and the moment of force relative to the point of application of the force. Each problem is presented in a separate block, which provides a step-by-step explanation of the solution process using appropriate formulas and methods.

The product design is made in a beautiful html format, which makes it convenient and enjoyable to read. There are also bright headings for each task, making it easy to navigate and find the information you need.

By purchasing this product, you receive useful and high-quality material that will help you better understand the material on linear and vector algebra, as well as prepare for an exam or test.

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