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Find vector \((\mathbf{a}-2 \mathbf{b}) \times \mathbf{c}, \quad\) where \(\mathbf{a}=\left|\begin{array}{rrr}{\mathbf{i}} & {\mathbf{j} \mathbf{k}} \\\ {2} & {-1} & {5} \\ {0} & {1} & {8}\end{array}\right|,\) \(\mathbf{b}=\left|\begin{array}{rrr}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\\ {0} & {1} & {1} \\ {2} & {-1} & {-2}\end{array}\right|, \quad\) and \(\mathbf{c}=\mathbf{i}+\mathbf{j}+\mathbf{k}.\)

Short Answer

Expert verified
The vector is \(-26\mathbf{i} + 17\mathbf{j} + 9\mathbf{k}\).

Step by step solution

01

Calculate Vector a

To compute \( \mathbf{a} \), use the determinant of the matrix given:\[ \mathbf{a} = \left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \{2} & {-1} & {5} \{0} & {1} & {8}\end{array}\right| \]Applying the determinant formula for a 3x3 matrix, we find:\[\mathbf{a} = \mathbf{i}(({-1})(8) - (5)(1)) - \mathbf{j}((2)(8) - (5)(0)) + \mathbf{k}((2)(1) - ((-1)(0))) \\mathbf{a} = \mathbf{i}(-8 - 5) - \mathbf{j}(16 - 0) + \mathbf{k}(2 - 0) \\mathbf{a} = -13\mathbf{i} - 16\mathbf{j} + 2\mathbf{k}\]
02

Calculate Vector b

Find \( \mathbf{b} \) using the determinant of its corresponding matrix:\[ \mathbf{b} = \left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \{0} & {1} & {1} \{2} & {-1} & {-2} \end{array}\right| \]Applying the determinant formula, we get:\[\mathbf{b} = \mathbf{i}((1)(-2) - (1)(-1)) - \mathbf{j}((0)(-2) - (1)(2)) + \mathbf{k}((0)(-1) - (1)(2)) \\mathbf{b} = \mathbf{i}(-2 + 1) - \mathbf{j}(0 - 2) + \mathbf{k}(0 - 2) \\mathbf{b} = -\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\]
03

Calculate Vector (a - 2b)

Calculate the vector \( \mathbf{a} - 2\mathbf{b} \):\[ 2\mathbf{b} = 2(-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = -2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k} \]Therefore,\[(\mathbf{a} - 2\mathbf{b}) = (-13\mathbf{i} - 16\mathbf{j} + 2\mathbf{k}) - (-2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}) \= (-13 + 2)\mathbf{i} + (-16 - 4)\mathbf{j} + (2 + 4)\mathbf{k} \= -11\mathbf{i} - 20\mathbf{j} + 6\mathbf{k}\]
04

Calculate Cross Product with c

Perform the cross product \((\mathbf{a} - 2\mathbf{b}) \times \mathbf{c}\):Given \( \mathbf{c} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), we set up the determinant:\[(\mathbf{a} - 2\mathbf{b}) \times \mathbf{c} = \left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \{-11} & {-20} & {6} \{1} & {1} & {1}\end{array}\right|\]Apply the determinant formula:\[= \mathbf{i}((-20)(1) - (6)(1)) - \mathbf{j}((-11)(1) - (6)(1)) + \mathbf{k}((-11)(1) - (-20)(1)) \= \mathbf{i}(-20 - 6) - \mathbf{j}(-11 - 6) + \mathbf{k}(-11 + 20) \= -26\mathbf{i} + 17\mathbf{j} + 9\mathbf{k}\]
05

Final Result

The resulting vector from the cross product \((\mathbf{a} - 2\mathbf{b}) \times \mathbf{c}\) is:\[-26\mathbf{i} + 17\mathbf{j} + 9\mathbf{k}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant of a matrix
The determinant of a matrix is a special number that can be calculated from a square matrix. It is useful for many applications in linear algebra, including solving systems of linear equations, finding the inverse of a matrix, and understanding vector spaces. The determinant of a 2x2 matrix \[ \begin{pmatrix} a & b \ c & d \end{pmatrix} \] is calculated as \( ad - bc \). For a 3x3 matrix, the determinant becomes a little more complex. Given a matrix:\[ \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \],its determinant is calculated using the formula:\[ a(ei - fh) - b(di - fg) + c(dh - eg) \]. The determinant helps us analyze linear transformations and space dimensions. In the context of 3D vectors, it can determine volume scaling when vectors represent geometric shapes like cubes or parallelepipeds.
Cross product
The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. The result is a vector that is perpendicular to the plane containing both original vectors. This operation is essential in physics and engineering for finding a vector normal to a surface.To find the cross product of vectors \( \mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} \) and \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), you can set up a determinant:\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \].The result is a new vector,\[ (u_2v_3 - u_3v_2)\mathbf{i} - (u_1v_3 - u_3v_1)\mathbf{j} + (u_1v_2 - u_2v_1)\mathbf{k} \].The magnitude of this resulting vector corresponds to the area of the parallelogram spanned by the original two vectors. Its direction is given by the right-hand rule.
Vector operations
Vector operations are fundamental actions you can perform with vectors that include addition, subtraction, scalar multiplication, and dot and cross products. They are crucial for vector calculus and help solve many physical and engineering problems.
  • Addition and Subtraction: Add or subtract corresponding components. If \( \mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} \) and \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), then \( \mathbf{u} + \mathbf{v} = (u_1+v_1)\mathbf{i} + (u_2+v_2)\mathbf{j} + (u_3+v_3)\mathbf{k} \).
  • Scalar Multiplication: Multiply each component by the scalar. If \( c \) is a scalar and \( \mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} \), then \( c \mathbf{u} = cu_1\mathbf{i} + cu_2\mathbf{j} + cu_3\mathbf{k} \).
  • Dot Product: Produces a scalar. Calculate as \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \).
These operations make it easy to manipulate vectors, solve projections, and measure magnitudes.
3D vectors
3D vectors add complexity by involving three-dimensional space, making them vital for representing real-world physical phenomena. A 3D vector is an ordered triplet \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z axes respectively.3D vectors are crucial for describing:
  • Forces and Velocities: In physics, these vectors define directions and magnitudes of forces, velocities, and accelerations.
  • Positions and Displacements: Represent the location of points in space or the space occupied by bodies.
  • Surfaces and Normals: Used in computer graphics and CAD to calculate normals for rendering and machining processes.
Understanding 3D vectors allows one to navigate and manipulate the space in modeling and simulations, providing a framework for analyzing real-life situations and solving complex problems.

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Most popular questions from this chapter

Determine whether the line of parametric equations \(x=1+2 t, y=-2 t, z=2+t, \quad t \in \mathbb{R}\) intersects the plane with equation \(3 x+4 y+6 z-7=0\) . If it does intersect, find the point of intersection.

Find the distance from point \(P(1,5,-4)\) to the plane of equation \(3 x-y+2 z-6=0\)

For the following exercises, find the equation of the plane with the given properties. The plane that passes through point \((4,7,-1)\) and has normal vector \(\mathbf{n}=\langle 3,4,2\rangle\)

[T]Let \(\mathbf{r}(t)=\left\langle t, 2 t^{2}, 4 t^{2}\right\rangle\) be the position vector of a particle at time \(t\) (in seconds), where \(t \in[0,10]\) (here the components of \(\mathbf{r}\) are expressed in centimeters). a. Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Round your answer to two decimal places. b. Use a CAS to visualize the path of the particle defined by the points \(\left(t, 2 t^{2}, 4 t^{2}\right), \quad\) where \(t \in[0,60] .\)

The force vector \(\mathbf{F}\) acting on a proton with an electric charge of \(1.6 \times 10^{-19} \mathrm{C}\) (in coulombs) moving in a magnetic field \(\mathbf{B}\) where the velocity vector \(\mathbf{v}\) is given by \(\mathbf{F}=1.6 \times 10^{-19}(\mathbf{v} \times \mathbf{B})\) (here, \(\mathbf{v}\) is expressed in meters per second, \(\mathbf{B}\) is in tesla \([\mathrm{T}],\) and \(\mathbf{F}\) is in newtons \([\mathrm{N}] )\) . Find the force that acts on a proton that moves in the \(x y\) -plane at velocity \(\mathbf{v}=10^{5} \mathbf{i}+10^{5} \mathbf{j}\) (in meters per second) in a magnetic field given by \(\mathbf{B}=0.3 \mathbf{j} .\)

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