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Chapter 14: Problem 12

If \(\boldsymbol{a}=3 \boldsymbol{i}+\boldsymbol{j}-2 \boldsymbol{k}\) and \(\boldsymbol{b}=13 \boldsymbol{i}-\boldsymbol{j}-\boldsymbol{k}\) find (a) \(a+b\) (b) \(b-3 a\) (c) \(|b|\) (e) \(|\boldsymbol{b}-\boldsymbol{a}|\) (d) \(\hat{a}\)

Short Answer

Expert verified
(a) \( 16\boldsymbol{i} - 3\boldsymbol{k} \); (b) \( 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k} \); (c) \( 13.08 \); (d) \( \hat{a} = \frac{3}{\sqrt{14}}\boldsymbol{i} + \frac{1}{\sqrt{14}}\boldsymbol{j} - \frac{2}{\sqrt{14}}\boldsymbol{k} \); (e) \( 10.25 \).

Step by step solution

01

Calculate \(a+b\)

Given \( \boldsymbol{a} = 3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k} \) and \( \boldsymbol{b} = 13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k} \), add the vectors:\[ \boldsymbol{a} + \boldsymbol{b} = (3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k}) + (13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k}) \]Combine like terms:\[ (3+13)\boldsymbol{i} + (1-1)\boldsymbol{j} + (-2-1)\boldsymbol{k} = 16\boldsymbol{i} - 3\boldsymbol{k} \]Thus, \( \boldsymbol{a} + \boldsymbol{b} = 16\boldsymbol{i} - 3\boldsymbol{k} \).
02

Calculate \( b-3a \)

Subtract \( 3 \boldsymbol{a} \) from \( \boldsymbol{b} \):\[ \boldsymbol{b} - 3\boldsymbol{a} = (13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k}) - 3(3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k}) \]Calculate \( 3\boldsymbol{a} \):\[ 3\boldsymbol{a} = 9\boldsymbol{i} + 3\boldsymbol{j} - 6\boldsymbol{k} \]Now perform the subtraction:\[ (13\boldsymbol{i} - 9\boldsymbol{i}) + (-\boldsymbol{j} - 3\boldsymbol{j}) + (-\boldsymbol{k} + 6\boldsymbol{k}) = 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k} \]Hence, \( \boldsymbol{b} - 3\boldsymbol{a} = 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k} \).
03

Calculate \(|b|\)

The magnitude of \( \boldsymbol{b} \) is found using:\[ |\boldsymbol{b}| = \sqrt{13^2 + (-1)^2 + (-1)^2} \]Simplify:\[ |\boldsymbol{b}| = \sqrt{169 + 1 + 1} = \sqrt{171} \]Thus, \(|\boldsymbol{b}| \approx 13.08 \).
04

Calculate \(|\boldsymbol{b} - \boldsymbol{a}|\)

First, find \( \boldsymbol{b} - \boldsymbol{a} \):\[ \boldsymbol{b} - \boldsymbol{a} = (13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k}) - (3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k}) \]Simplify the subtraction:\[ (13-3)\boldsymbol{i} + (-1-1)\boldsymbol{j} + (-1+2)\boldsymbol{k} = 10\boldsymbol{i} - 2\boldsymbol{j} + \boldsymbol{k} \]Then, calculate the magnitude:\[ |\boldsymbol{b} - \boldsymbol{a}| = \sqrt{10^2 + (-2)^2 + 1^2} = \sqrt{100 + 4 + 1} = \sqrt{105} \]Thus, \(|\boldsymbol{b} - \boldsymbol{a}| \approx 10.25\).
05

Calculate \( \hat{a} \)

The unit vector \( \hat{a} \) is found by dividing \( \boldsymbol{a} \) by its magnitude:\[ \hat{a} = \frac{\boldsymbol{a}}{|\boldsymbol{a}|} \]First, determine \(|\boldsymbol{a}|\):\[ |\boldsymbol{a}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9+1+4} = \sqrt{14} \]Now calculate \( \hat{a} \):\[ \hat{a} = \frac{3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k}}{\sqrt{14}} = \frac{3}{\sqrt{14}}\boldsymbol{i} + \frac{1}{\sqrt{14}}\boldsymbol{j} - \frac{2}{\sqrt{14}}\boldsymbol{k} \]

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Key Concepts

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

Vector Addition
When adding two vectors, the process involves combining like components. A vector is represented by its components along the unit vectors: usually denoted as \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). Each component corresponds to the movement or scaling in the respective axis (x, y, and z).
  • Consider two vectors \( \boldsymbol{a} = 3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k} \) and \( \boldsymbol{b} = 13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k} \).
  • To add these vectors, simply add their corresponding components:\[ \boldsymbol{a} + \boldsymbol{b} = (3 + 13)\boldsymbol{i} + (1 - 1)\boldsymbol{j} + (-2 - 1)\boldsymbol{k} \]
  • This results in the vector \( 16\boldsymbol{i} - 3\boldsymbol{k} \).
Vector addition helps in determining the resultant vector that represents the cumulative effect of multiple distinct vectors.
Magnitude of a Vector
The magnitude of a vector, often termed as the vector's length or size, measures how long the vector is. It is a non-negative scalar. To find the magnitude, you use the formula derived from the Pythagorean theorem.
  • For a vector \( \boldsymbol{b} = 13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k} \), the magnitude is calculated as:\[|\boldsymbol{b}| = \sqrt{13^2 + (-1)^2 + (-1)^2} = \sqrt{171}\]
  • This is approximately \( 13.08 \).
Understanding the magnitude is crucial for comparing vectors and determining distances in fields like physics and engineering.
Unit Vector
A unit vector is a vector that has a magnitude of 1. It indicates direction without magnitude. Finding a unit vector in the direction of a given vector \( \boldsymbol{a} = 3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k} \) requires scaling down the vector by its magnitude.
  • First, calculate the magnitude of \( \boldsymbol{a} \):\[|\boldsymbol{a}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{14}\]
  • Then, divide each component of \( \boldsymbol{a} \) by \( \sqrt{14} \):\[\hat{a} = \frac{3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k}}{\sqrt{14}}\]
A unit vector is particularly useful in applications requiring directional information, such as assigning a precise heading to an object in motion.
Vector Subtraction
Vector subtraction is similar to addition but involves combining components by their difference. It is essentially adding a vector in the opposite direction.
  • Given vectors \( \boldsymbol{b} \) and \( 3\boldsymbol{a} \), first calculate \( 3\boldsymbol{a} \):\[3\boldsymbol{a} = 9\boldsymbol{i} + 3\boldsymbol{j} - 6\boldsymbol{k}\]
  • Then subtract \( 3\boldsymbol{a} \) from \( \boldsymbol{b} \):\[\boldsymbol{b} - 3\boldsymbol{a} = (13\boldsymbol{i} - 9\boldsymbol{i}) + (-1\boldsymbol{j} - 3\boldsymbol{j}) + (-1\boldsymbol{k} + 6\boldsymbol{k}) = 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k}\]
Vector subtraction can help in determining relative position change, force balance, and other fields requiring differential analysis.

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Most popular questions from this chapter

Find the projection of the vector \(6 \boldsymbol{i}+\boldsymbol{j}+5 \boldsymbol{k}\) onto the vector \(\boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}\).

Find the modulus of each of the following vectors: (a) \(r=7 i+3 j\) (b) \(r=17 i\) (c) \(r=2 i-3 j\) (d) \(\boldsymbol{r}=-3 \boldsymbol{j}\) (e) \(\boldsymbol{r}=a \boldsymbol{i}+b \boldsymbol{j}\) (f) \(\boldsymbol{r}=a \boldsymbol{i}-b \boldsymbol{j}\)

Show on a diagram three arbitrary vectors \(p, q\) and \(r\). Using the triangle law of addition verify the associativity rule $$ p+(q+r)=(p+q)+r $$

Find the vector equation of the line passing through \((9,1,2)\) and which is parallel to the vector \((1,1,1)\).

Use the definition of the scalar product to show that, if two vectors are perpendicular, their scalar product is zero.
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