/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Given that \(a=3 i-j-4 k, b=-2 i... [FREE SOLUTION] | 91Ó°ÊÓ

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Given that \(a=3 i-j-4 k, b=-2 i+4 j-3 k\) and \(c=i+2 j-k\), find (a) the magnitude of the vector \(a+b+c\); (b) a unit vector parallel to \(3 a-2 b+4 c\); (c) the angles between the vectors \(a\) and \(b\) and between \(b\) and \(c\); (d) the position vector of the centre of mass of particles of masses 1,2 and 3 placed at points \(A\), \(\mathrm{B}\) and \(\mathrm{C}\) with position vectors \(\boldsymbol{a}, \boldsymbol{b}\) and \(\boldsymbol{c}\) respectively.

Short Answer

Expert verified
(a) \( \sqrt{93} \); (b) \( \frac{17}{\sqrt{398}}i - \frac{3}{\sqrt{398}}j - \frac{10}{\sqrt{398}}k \); (c) \( \cos\theta_{ab} = \frac{2}{\sqrt{754}} \); (d) \( \frac{1}{3}i + \frac{13}{6}j - \frac{13}{6}k \).

Step by step solution

01

Find Vector a+b+c

To find the vector sum of \(a\), \(b\), and \(c\), add their corresponding components. \(a = 3i - j - 4k\), \(b = -2i + 4j - 3k\), and \(c = i + 2j - k\). Add them to obtain \(a + b + c = (3 - 2 + 1)i + (-1 + 4 + 2)j + (-4 - 3 - 1)k\). Hence, \(a + b + c = 2i + 5j - 8k\).
02

Calculate Magnitude of a+b+c

Use the formula \( \|a+b+c\| = \sqrt{a^2 + b^2 + c^2} \) for the vector \(2i + 5j - 8k\). Here, \(a=2\), \(b=5\), and \(c=-8\). Calculate: \( \|a+b+c\| = \sqrt{2^2 + 5^2 + (-8)^2} = \sqrt{4 + 25 + 64} = \sqrt{93} \). Thus, the magnitude is \( \sqrt{93} \).
03

Find Vector 3a - 2b + 4c

Substitute the given vectors and calculate. \(3a = 9i - 3j - 12k\), \(-2b = 4i - 8j + 6k\), \(4c = 4i + 8j - 4k\). Combine them: \(3a - 2b + 4c = (9 + 4 + 4)i + (-3 - 8 + 8)j + (-12 + 6 - 4)k = 17i - 3j - 10k\).
04

Calculate Unit Vector for 3a - 2b + 4c

The unit vector in the direction of \(3a - 2b + 4c = 17i - 3j - 10k\) is given by \( \frac{1}{\text{magnitude}} (17i - 3j - 10k) \). First, find the magnitude: \(\sqrt{17^2 + (-3)^2 + (-10)^2} = \sqrt{389 + 9 + 100} = \sqrt{398}\). \(\therefore\) the unit vector is \( \frac{1}{\sqrt{398}} (17i - 3j - 10k) \).
05

Find Dot Product for Angles

The cosine of the angle between two vectors \(u\) and \(v\) is given by \(\cos\theta = \frac{u \cdot v}{\|u\| \cdot \|v\|}\). Calculate \(a \cdot b = (3)(-2) + (-1)(4) + (-4)(-3) = -6 - 4 + 12 = 2\). For \(a\), \(\|a\| = \sqrt{26}\) and for \(b\), \(\|b\| = \sqrt{29}\). Thus, \(\cos\theta_{ab} = \frac{2}{\sqrt{754}}\). Repeat for vectors \(b\) and \(c\).
06

Calculate Center of Mass Position Vector

For masses \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = 3\) with position vectors \(a\), \(b\), and \(c\) respectively, use the formula for the center of mass position vector: \(r_{cm} = \frac{m_1a + m_2b + m_3c}{m_1 + m_2 + m_3}\). Calculate: \(r_{cm} = \frac{1(3i-j-4k) + 2(-2i+4j-3k) + 3(i+2j-k)}{6} = \frac{(3i-j-4k) + (-4i+8j-6k) + (3i+6j-3k)}{6}\). Result: \(r_{cm} = \frac{2i + 13j - 13k}{6} = \frac{1}{3}i + \frac{13}{6}j - \frac{13}{6}k\).

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Key Concepts

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

Vector Magnitude
Finding the magnitude of a vector is like measuring its length. Imagine you're holding a stick in space—the magnitude tells you how long the stick is, but not its direction. In mathematical terms, the magnitude of a vector is computed using the square root of the sum of the squares of its components. For example, given a vector \(v = xi + yj + zk\), the magnitude \(\|v\|\) is calculated by the formula:\[\|v\| = \sqrt{x^2 + y^2 + z^2}\]This formula derives from the Pythagorean theorem in three dimensions. It finds applications in physics and engineering wherever vector quantities are analyzed. Calculating the magnitude can help determine the overall strength or intensity of a vector quantity, such as force or velocity.
Understanding vector magnitude is key in determining distances and sizes within vector calculus.
Unit Vector
Unit vectors are incredibly useful in vector operations as they provide direction while having a magnitude of exactly one. This makes them scale-neutral and ideal for defining directions.Creating a unit vector involves taking an existing vector and dividing it by its magnitude. For a vector \(v = xi + yj + zk\), you'll first compute its magnitude \(\|v\|\) and then form the unit vector \(\hat{v}\) as:\[\hat{v} = \frac{v}{\|v\|} = \frac{xi + yj + zk}{\sqrt{x^2 + y^2 + z^2}}\]Unit vectors are foundational in defining coordinates such as \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) in 3D space. By normalizing vectors to unit length, they become easier to work with in calculations involving vector directions.
Unit vectors are like directional arrows in space, making them indispensable in navigation and physics.
Dot Product
The dot product, also known as the scalar product, is a method of multiplying two vectors that results in a scalar quantity. This scalar can give you information about the angle between the vectors or the work done in physics. To compute the dot product of two vectors \(u = xi + yj + zk\) and \(v = ai + bj + ck\), use the formula:\[u \cdot v = x a + y b + z c\]This product can be interpreted geometrically as:\[u \cdot v = \|u\| \times \|v\| \times \cos(\theta)\]Where \(\theta\) is the angle between the two vectors. This relationship helps in finding the angle if you know the vectors. The dot product can identify whether vectors are orthogonal (if the dot product is zero, the vectors are perpendicular) or gauge their alignment.
Understanding dot products is critical when working with projections and calculating angles.
Center of Mass
The center of mass is a critical concept in physics and engineering, representing the average position of all the mass in a system. For discrete particles or masses, the center of mass is the point at which you could support the object, and it would balance. Mathematically, it's defined by the weighted average of their positions.Assume particles with masses \(m_1, m_2, ... m_n\) are placed at position vectors \(r_1, r_2, ... r_n\). The center of mass position vector \(r_{cm}\) is given as:\[r_{cm} = \frac{m_1r_1 + m_2r_2 + ... + m_nr_n}{m_1+m_2+...+m_n}\]This formula provides a means to calculate where the overall mass of the system can be considered to be concentrated. The center of mass is pivotal in understanding the dynamics of a system, whether it's a solid body in space or a collection of particles.
Grasping the concept of center of mass is essential in fields such as mechanical design and astrophysics.

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Most popular questions from this chapter

A particle of mass \(m\), charge \(e\) and moving with velocity \(\nu\) in a magnetic field of strength \(H\) is known to have acceleration $$ \frac{e}{m c}(\mathbf{v} \times \boldsymbol{H}) $$ where \(c\) is the speed of light. Show that the component of acceleration parallel to \(\boldsymbol{H}\) is zero.

Given that \(u=(4,0,-2), v=(3,1,-1)\), \(w=(2,1,6)\) and \(s=(1,4,1)\), evaluate (a) \(\boldsymbol{u} \cdot \boldsymbol{v}\) (b) \(v \cdot s\) (c) \(\hat{w}\) (d) \((v \cdot s) \hat{u}\) (e) \((\boldsymbol{u} \cdot \boldsymbol{w})(\boldsymbol{v} \cdot \boldsymbol{s})\) (f) \((u \cdot i) v+(w \cdot s) k\)

Given the line with parametric equation $$ \boldsymbol{r}=\boldsymbol{a}+\lambda d $$ show that the perpendicular distance \(p\) from the origin to this line can take either of the forms (i) \(p=\frac{|a \times d|}{|d|}\) (ii) \(p=\left|a-\frac{a \cdot d}{d \cdot d} d\right|\) Find the parametric equation of the straight line through the points \(\mathrm{A}(1,0,2)\) and \(\mathrm{B}(2,3,0)\) and determine (a) the length of the perpendicular from the origin to the line; (b) the point at which the line intersects the \(y-z\) plane; (c) the coordinates of the foot of the perpendicular to the line from the point \((1,1,1)\)

Given the triangle \(\mathrm{OAB}\), where \(\mathrm{O}\) is the origin, and denoting the midpoints of the opposite sides as \(\mathrm{O}^{\prime}, \mathrm{A}^{\prime}\) and \(\mathrm{B}^{\prime}\), show vectorially that the lines \(\mathrm{OO}^{\prime}, \mathrm{AA}^{\prime}\) and \(\mathrm{BB}^{\prime}\) meet at a point. (Note that this is the result that the medians of a triangle meet at the centroid.)

Show that the equation of the plane through the points \(P_{1}, P_{2}\) and \(P_{3}\) with position vectors \(\left.\boldsymbol{r}_{1}, \boldsymbol{r}_{2}\right\\}\) and \(r_{3}\) respectively takes the form $$ \boldsymbol{r} \cdot\left[\left(\boldsymbol{r}_{1} \times \boldsymbol{r}_{2}\right)+\left(\boldsymbol{r}_{2} \times \boldsymbol{r}_{3}\right)+\left(\boldsymbol{r}_{3} \times \boldsymbol{r}_{1}\right)\right]=\boldsymbol{r}_{1} \cdot\left(\boldsymbol{r}_{2} \times \boldsymbol{r}_{3}\right) $$

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