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Given \(\mathbf{M}=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\mathbf{N}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}},\) calculate the vector product \(\mathbf{M} \times \mathbf{N} .\)

Short Answer

Expert verified
The vector product of \(\mathbf{M}\) and \(\mathbf{N}\) is \(-7\hat{\mathbf{i}} + 20\hat{\mathbf{j}} -10\hat{\mathbf{k}}\).

Step by step solution

01

Write down formula

The formula for the cross product of two vectors is given by: \(\mathbf{A} \times \mathbf{B} = (A_yB_z - A_zB_y)\mathbf{i} - (A_xB_z - A_zB_x)\mathbf{j} + (A_xB_y - A_yB_x)\mathbf{k}\)
02

Substitute the values into the formula

Substitute the values of vector M and N into the cross product formula to calculate the components of the cross product vector: \(M_x = 6, M_y = 2, M_z = -1, N_x = 2, N_y = -1, N_z = -3\)
03

Calculate the i component

Using the cross product formula, calculate the i component of the resulting vector, this can be done as follows: \((M_yN_z - M_zN_y)i = (2*-3 - (-1)*-1)i = -6 - 1 = -7i\)
04

Calculate the j component

Similarly, calculate the j component of the resulting vector as follows: \(-(M_xN_z - M_zN_x)j = -(6*-3 - (-1)*2)j = -(-18 - 2)j = -(-20)j = 20j\)
05

Calculate the k component

Finally, calculate the k component of the resulting vector as follows: \((M_xN_y - M_yN_x)k = (6*-1 - 2*2)k = -6 - 4 = -10k\)
06

Combine results

Combine the i, j, and k components calculated in steps 3, 4 and 5 to find the vector product \( M \times N\): \(-7i + 20j -10k\)

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Key Concepts

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

Cross Product
The cross product, also known as the vector product, is a fundamental operation in vector mathematics. It is used to calculate a vector that is orthogonal to two given vectors. This is especially useful in physics and engineering to find perpendicular directions.To define the cross product, consider two vectors \( \mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}} \) and \( \mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}} \). The cross product between these vectors, \( \mathbf{A} \times \mathbf{B} \), results in another vector calculated using:
  • \( i \)-component: \((A_yB_z - A_zB_y)\hat{\mathbf{i}}\)
  • \( j \)-component: \(-(A_xB_z - A_zB_x)\hat{\mathbf{j}}\)
  • \( k \)-component: \((A_xB_y - A_yB_x)\hat{\mathbf{k}}\)
Each component is calculated through determinant-like structures, succinctly capturing the geometric intricacy of how the vectors align in three-dimensional space. Remember, the cross product will always be perpendicular to both involved vectors.
Vector Mathematics
In vector mathematics, understanding operations and their geometric interpretations is crucial. Vectors are not merely numbers but quantities that possess both magnitude and direction. This enables a wide range of applications in different fields, like physics, engineering, and computer graphics.Key operations include:
  • Addition and Subtraction: Performed component-wise, influencing magnitude and direction together.
  • Dot Product: Measures vector alignment, outputting a scalar, and essential for understanding angles between vectors.
  • Cross Product: Yields a vector, vital for determining perpendicularity as demonstrated in our problem with vectors \( \mathbf{M} \) and \( \mathbf{N} \).
For effective vector manipulation, learning to switch smoothly between these operations and visualizing their results in physical terms can greatly enhance comprehension and problem-solving efficiency.
Physics Problem Solving
In physics problem-solving, vectors provide an effective way to break down forces, velocities, and other physical quantities. The cross product is particularly useful when dealing with rotational forces (torques) or when calculating the magnetic force on a charge moving in a magnetic field.Physics problems often require:
  • Identifying Vector Components: Breaking vectors into \( i \), \( j \), and \( k \) components helps simplify complex problems into manageable parts.
  • Utilizing Cross Product: Essential for finding resultant vectors that need directional consideration, like the resultant of forces in mechanics.
  • Applying Conceptual Understandings: Integrating mathematical results with physical insights, allowing for predictions and solutions that align with real-world behaviors.
Approaching physics problems with a strong foundation in vector operations supports the intuitive understanding necessary to tackle real-world challenges effectively, as seen in our cross product calculation for vectors \( \mathbf{M} \) and \( \mathbf{N} \).

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Most popular questions from this chapter

A spacecraft is in empty space. It carries on board a gyroscope with a moment of inertia of \(I_{g}=20.0 \mathrm{kg} \cdot \mathrm{m}^{2}\) about the axis of the gyroscope. The moment of inertia of the spacecraft around the same axis is \(I_{s}=5.00 \times 10^{5} \mathrm{kg} \cdot \mathrm{m}^{2}\) Neither the spacecraft nor the gyroscope is originally rotating. The gyroscope can be powered up in a negligible period of time to an angular speed of \(100 \mathrm{s}^{-1} .\) If the orientation of the spacecraft is to be changed by \(30.0^{\circ},\) for how long should the gyroscope be operated?

A uniform rod of mass \(100 \mathrm{g}\) and length \(50.0 \mathrm{cm}\) rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small beads, each of mass \(30.0 \mathrm{g},\) are mounted on the rod so that they are able to slide without friction along its length. Initially the beads are held by catches at positions \(10.0 \mathrm{cm}\) on each side of center, at which time the system rotates at an angular speed of \(20.0 \mathrm{rad} / \mathrm{s} .\) Suddenly, the catches are released and the small beads slide outward along the rod. (a) Find the angular speed of the system at the instant the beads reach the ends of the rod. (b) What if the beads fly off the ends? What is the angular speed of the rod after this occurs?

A student sits on a freely rotating stool holding two weights, each of mass \(3.00 \mathrm{kg}\) (Figure \(\mathrm{P} 11.30\) ). When his arms are extended horizontally, the weights are \(1.00 \mathrm{m}\) from the axis of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is \(3.00 \mathrm{kg} \cdot \mathrm{m}^{2}\) and is assumed to be constant. The student pulls the weights inward horizontally to a position \(0.300 \mathrm{m}\) from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the weights inward.

Two vectors are given by \(\mathbf{A}=-3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\) and \(\mathbf{B}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} .\) Find \((\mathbf{a}) \mathbf{A} \times \mathbf{B}\) and \((\mathbf{b})\) the angle between A and B.

The vectors \(42.0 \mathrm{cm}\) at \(15.0^{\circ}\) and \(23.0 \mathrm{cm}\) at \(65.0^{\circ}\) both start from the origin. Both angles are measured counterclockwise from the \(x\) axis. The vectors form two sides of a parallelogram. (a) Find the area of the parallelogram. (b) Find the length of its longer diagonal.

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