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Chapter 9: Problem 8

Let \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} .\) Let \(\overrightarrow{\mathbf{c}}=\) \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\). ( \(a\) ) Find \(\mathbf{c}\), expressed in unit vector notation. ( \(b\) ) Find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\).

Short Answer

Expert verified
The resultant vector '\(\overrightarrow{\mathbf{c}}\)' after cross product operation is \((-11)\hat{\mathbf{i}} -2\hat{\mathbf{j}} + 10\hat{\mathbf{k}}\). The angle between vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is around 55.41 degrees.

Step by step solution

01

Cross product of vector 'a' and 'b'

The cross product of two vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is given by \(c_x = a_y*b_z - a_z*b_y\), \(c_y = a_z*b_x - a_x*b_z\), \(c_z = a_x*b_y - a_y*b_x\). Here, \(c_x\), \(c_y\), \(c_z\) are components of the resulted vector '\(\overrightarrow{\mathbf{c}}\)' in respective directions. By plugging in the given vectors, we get \(\overrightarrow{\mathbf{c}}=(-9 - 2)\hat{\mathbf{i}} + (6-8)\hat{\mathbf{j}} + (4 + 6)\hat{\mathbf{k}}\).
02

Calculate resultant vector '\(\overrightarrow{\mathbf{c}}\)'

This step involves simple arithmetic calculation for each component of '\(\overrightarrow{\mathbf{c}}\)'. We get \(\overrightarrow{\mathbf{c}}=(-11)\hat{\mathbf{i}} + (-2)\hat{\mathbf{j}} + 10\hat{\mathbf{k}}\).
03

Calculate dot product and Magnitude of vectors 'a' and 'b'

To find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\), we need to calculate dot product and magnitude. The dot product of two vectors is given by \(a_x*b_x + a_y*b_y + a_z*b_z\), and the magnitude of a vector is given by \(\sqrt{{x^2 + y^2 + z^2}}\). Thus, for our vectors, \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = (2*4)+(-3*-2)+(1*-3) = 8 + 6 -3 = 11\), also \(\|\overrightarrow{\mathbf{a}}\| = \sqrt{{2^2 + (-3)^2 + 1^2}} = \sqrt{4 + 9 +1} = \sqrt{14}\) and \(\|\overrightarrow{\mathbf{b}}\| = \sqrt{{4^2 + (-2)^2 + (-3)^2}} = \sqrt{16+4+9} = \sqrt{29}\).
04

Calculate the angle between vectors

The angle \(\theta\) between vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) can be calculated as \(\cos(\theta) = \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}} {\|\overrightarrow{\mathbf{a}}\|\*\|\overrightarrow{\mathbf{b}}\|}\). Substituting the calculated values, we get \(\cos(\theta) = \frac{11}{\sqrt{14} * \sqrt{29}} = 0.5785\) and \(\theta = \cos^{-1}(0.5785) = 55.41^\circ\).

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Most popular questions from this chapter

A parked automobile of mass \(1360 \mathrm{~kg}\) has a wheel base (distance between front and rear axles) of \(305 \mathrm{~cm}\). Its center of gravity is located \(178 \mathrm{~cm}\) behind the front axle. Determine \((a)\) the upward force exerted by the level ground on each of the front wheels (assumed the same) and \((b)\) the upward force exerted by the level ground on each of the rear wheels (assumed the same).

Vector \(\overrightarrow{\mathbf{a}}\) has magnitude \(3.20\) units and lies in the \(y z\) plane \(63.0^{\circ}\) from the \(+y\) axis with a positive \(z\) component. Vector \(\overrightarrow{\mathbf{b}}\) has magnitude \(1.40\) units and lies in the \(x z\) plane \(48.0^{\circ}\) from the \(+x\) axis with a positive \(z\) component. Find \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\).

Two vectors \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\) lie in the \(x y\) plane. Their magnitudes are \(r=4.5\) units and \(s=7.3\) units. Their directions are, respectively, \(320^{\circ}\) and \(85^{\circ}\) measured counterclockwise from the positive \(x\) axis. Find the magnitude and the direction of \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}\)

Two identical blocks, each of mass \(M\), are connected by a light string over a frictionless pulley of radius \(R\) and rotational inertia \(I\) (Fig. 9-55). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and the sliding block. When this system is released, it is found that the pulley turns through an angle \(\theta\) in time \(t\) and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? ( \(b\) ) What is the acceleration of the two blocks? ( \(c\) ) What are the tensions in the upper and lower sections of the string? All answers are to be expressed in terms of \(M, I, R, \theta, g\), and \(t\).

Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the 20 -cm mark.
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