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Chapter 4: Problem 44

Given the vectors \(\boldsymbol{a}=(-3,-1,-2)\) and \(\boldsymbol{b}=\) \((2,3,1)\), find \(|\boldsymbol{a} \times \boldsymbol{b}|\) and \((\boldsymbol{a}+2 \boldsymbol{b}) \times(2 \boldsymbol{a}-\boldsymbol{b})\)

Short Answer

Expert verified
\(|\boldsymbol{a} \times \boldsymbol{b}| = 5\sqrt{3}\); \((\boldsymbol{a}+2\boldsymbol{b}) \times(2\boldsymbol{a}-\boldsymbol{b}) \) has magnitude \(25\sqrt{3}\).

Step by step solution

01

Calculate the Cross Product \( \boldsymbol{a} \times \boldsymbol{b} \)

The cross product \( \boldsymbol{a} \times \boldsymbol{b} \) for vectors \( \boldsymbol{a}=(-3,-1,-2) \) and \( \boldsymbol{b}=(2,3,1) \) is determined using the determinant:\[\boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \ -3 & -1 & -2 \ 2 & 3 & 1 \end{vmatrix}\]Calculate the determinant:\[\boldsymbol{a} \times \boldsymbol{b} = \boldsymbol{i}((-1)(1) - (-2)(3)) - \boldsymbol{j}((-3)(1) - (-2)(2)) + \boldsymbol{k}((-3)(3) - (-1)(2))\]Simplifying gives:\[\boldsymbol{a} \times \boldsymbol{b} = \boldsymbol{i}(-1 + 6) - \boldsymbol{j}(-3 + 4) + \boldsymbol{k}(-9 + 2)\]\[\boldsymbol{a} \times \boldsymbol{b} = 5\boldsymbol{i} + 1\boldsymbol{j} - 7\boldsymbol{k}\]Thus, \( \boldsymbol{a} \times \boldsymbol{b} = (5, 1, -7) \).
02

Compute Magnitude of \( \boldsymbol{a} \times \boldsymbol{b} \)

The magnitude of the cross product \( \boldsymbol{a} \times \boldsymbol{b} = (5, 1, -7) \) is calculated as follows:\[|\boldsymbol{a} \times \boldsymbol{b}| = \sqrt{5^2 + 1^2 + (-7)^2}\]\[|\boldsymbol{a} \times \boldsymbol{b}| = \sqrt{25 + 1 + 49}\]\[|\boldsymbol{a} \times \boldsymbol{b}| = \sqrt{75}\]\[|\boldsymbol{a} \times \boldsymbol{b}| = 5\sqrt{3}\]
03

Calculate the Vector \( \boldsymbol{a} + 2\boldsymbol{b} \)

Calculate \( \boldsymbol{a} + 2\boldsymbol{b} \) where \( \boldsymbol{a}=(-3, -1, -2) \) and \( \boldsymbol{b}=(2, 3, 1) \):\[\boldsymbol{b} = 2 \times (2, 3, 1) = (4, 6, 2)\]\[\boldsymbol{a} + 2\boldsymbol{b} = (-3, -1, -2) + (4, 6, 2) = (1, 5, 0)\]
04

Calculate the Vector \( 2\boldsymbol{a} - \boldsymbol{b} \)

Calculate \( 2\boldsymbol{a} - \boldsymbol{b} \) using \( \boldsymbol{a}=(-3, -1, -2) \) and \( \boldsymbol{b}=(2, 3, 1) \):\[2\boldsymbol{a} = 2 \times (-3, -1, -2) = (-6, -2, -4)\]\[2\boldsymbol{a} - \boldsymbol{b} = (-6, -2, -4) - (2, 3, 1) = (-8, -5, -5)\]
05

Calculate Cross Product \( (\boldsymbol{a} + 2\boldsymbol{b}) \times (2\boldsymbol{a} - \boldsymbol{b}) \)

Using vectors from Step 3 \((1, 5, 0)\) and Step 4 \((-8, -5, -5)\):\[\boldsymbol{a} + 2\boldsymbol{b} = (1, 5, 0)\]\[2\boldsymbol{a} - \boldsymbol{b} = (-8, -5, -5)\]Calculate the determinant:\[\begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \ 1 & 5 & 0 \ -8 & -5 & -5 \end{vmatrix}\]\[= \boldsymbol{i}(5 \cdot -5 - 0) - \boldsymbol{j}(1 \cdot -5 - 0) + \boldsymbol{k}(1 \cdot -5 - 5 \cdot -8)\]Simplifying gives:\[= \boldsymbol{i}(-25) - \boldsymbol{j}(-5) + \boldsymbol{k}(-5 + 40)\]\[= -25\boldsymbol{i} + 5\boldsymbol{j} + 35\boldsymbol{k}\]
06

Find the Magnitude of Result from Step 5

Calculate the magnitude of the vector \((-25, 5, 35)\):\[|(-25, 5, 35)| = \sqrt{(-25)^2 + 5^2 + 35^2}\]\[= \sqrt{625 + 25 + 1225}\]\[= \sqrt{1875}\]\[= 25\sqrt{3}\]

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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 of two vectors is a vector that is perpendicular to both of the original vectors. This operation is crucial in 3D vector algebra. It is often used to find a vector perpendicular to a plane defined by two other vectors.
To calculate the cross product of two vectors, you'll use a determinant. For example, with vectors \( \boldsymbol{a}=(-3,-1,-2) \) and \( \boldsymbol{b}=(2,3,1) \), the cross product \( \boldsymbol{a} \times \boldsymbol{b} \) is calculated using:
  • The determinant expression for cross product: \[\boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \-3 & -1 & -2 \2 & 3 & 1 \end{vmatrix}\]
  • Solve the determinant by following the rule: \( \boldsymbol{i}((-1)(1) - (-2)(3)) - \boldsymbol{j}((-3)(1) - (-2)(2)) + \boldsymbol{k}((-3)(3) - (-1)(2)) \).
  • Simplifying this gives the vector \((5, 1, -7)\), which is the cross product of \( \boldsymbol{a} \) and \( \boldsymbol{b} \).
In summary, the cross product provides us a new vector that helps identify the orientation and perpendicularity of vectors in space.
Magnitude of a Vector
The magnitude of a vector, often thought of as the length or size, is an important concept in vector calculus. It offers insights about how large a vector is relative to the origin.
To find the magnitude of a vector given by its components \( (x, y, z) \), apply the following formula:
  • Magnitude formula: \[|\boldsymbol{v}| = \sqrt{x^2 + y^2 + z^2}\]
  • For example, using our cross product \( \boldsymbol{a} \times \boldsymbol{b} = (5, 1, -7) \), the magnitude is: \[|\boldsymbol{a} \times \boldsymbol{b}| = \sqrt{5^2 + 1^2 + (-7)^2}\]
  • On solving, this becomes \( \sqrt{25 + 1 + 49} = \sqrt{75} \), which simplifies to \( 5\sqrt{3} \).
Thus, the magnitude of a vector gives us the scalar representation of the vector's length, which is useful for understanding scale in vector-based problems.
Determinant
The determinant is a special number that can be calculated from a square matrix and is essential for finding solutions in linear algebra, especially in vector calculus when dealing with cross products.
For a 3x3 matrix of the form:
  • Matrix format: \[\begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix}\]
  • The determinant is computed as: \[\boldsymbol{i}(b_1c_2 - b_2c_1) - \boldsymbol{j}(a_1c_2 - a_2c_1) + \boldsymbol{k}(a_1b_2 - a_2b_1)\]
  • This calculation helps in deriving the cross product: \[\boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \-3 & -1 & -2 \2 & 3 & 1 \end{vmatrix} = 5\boldsymbol{i} + 1\boldsymbol{j} - 7\boldsymbol{k}\]
The determinant, thus, plays a pivotal role in vector calculus by facilitating complex calculations like identifying cross products and understanding vector interactions within a space.

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Most popular questions from this chapter

The vector \(\overrightarrow{\mathrm{OP}}\) makes an angle of \(60^{\circ}\) with the positive \(x\) axis and \(45^{\circ}\) with the positive \(y\) axis. Find the possible angles that the vector can make with the \(z\) axis.

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

Given \(|\boldsymbol{a}|=3,|\boldsymbol{b}|=2\) and \(\boldsymbol{a} \cdot \boldsymbol{b}=5\) find \(|\boldsymbol{a}+2 \boldsymbol{b}|\) and \(|3 \boldsymbol{a}-\boldsymbol{b}| .\) Find the angle between the vectors \(a+2 b\) and \(3 \boldsymbol{a}-\boldsymbol{b} .\)

Find the equation of the plane with perpendicular \(\boldsymbol{n}=(1,-1,1)\) that passes through the point with position vector \((2,3,3)\). Show that the line with equation \(\boldsymbol{r}=(-1,-1,2)+\boldsymbol{t}(2,0,-2)\) lies in this plane.

Find the constant \(\lambda\) such that the three vectors \((3,2,-1),(1,-1,3)\) and \((2,-3, \lambda)\) are coplanar.
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