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Chapter 9: Problem 10

For what value(s) of \(b\) are the vectors \((-6, b)\) and \(\left(b, b^{2}\right)\) perpendicular?

Short Answer

Expert verified
The values of \(b\) are 0, \(\sqrt{6}\), and \(-\sqrt{6}\).

Step by step solution

01

Understanding Vector Perpendicularity

Two vectors are perpendicular if their dot product is zero. Given vectors \((-6, b)\) and \( (b, b^2)\), we can use the dot product formula to find the condition for perpendicularity.
02

Writing the Dot Product

The dot product of vectors \((-6, b)\) and \( (b, b^2)\) is calculated as: \[-6 \times b + b \times b^2\] which simplifies to: \[-6b + b^3\] Set this expression to zero for the vectors to be perpendicular.
03

Setting Up the Equation

From the dot product, we have the equation: \[-6b + b^3 = 0\]to solve for \(b\).
04

Factoring the Expression

Factor the equation \{-6b + b^3 = 0\}:\[b(b^2 - 6) = 0\]This gives us two possible solutions: \(b = 0\) or \(b^2 - 6 = 0\).
05

Solving for b

The first solution is straightforward: \(b = 0\).For the second solution, solve \(b^2 - 6 = 0\) by adding 6 to both sides, resulting in \(b^2 = 6\). Taking the square root of both sides, we get \(b = \pm\sqrt{6}\).
06

Listing All Solutions

The values of \(b\) for which the vectors are perpendicular are: \(b = 0\), \(b = \sqrt{6}\), and \(b = -\sqrt{6}\).

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Key Concepts

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

Perpendicularity and Vectors
Perpendicularity in geometry and vector mathematics signifies a special relationship between two vectors. When two vectors are perpendicular, they meet at a 90-degree angle. This is an important concept because perpendicular vectors have properties that simplify many mathematical calculations.

To check for perpendicularity between two vectors, say \(\mathbf{a}\) and \(\mathbf{b}\), we examine their dot product. If the dot product equals zero, the vectors are perpendicular.

Understanding this property helps us solve problems where we want to determine if vectors align perpendicularly. This approach was used in the exercise when trying to find the value of \(b\) that makes vectors \((-6, b)\) and \((b, b^2)\) perpendicular."},{

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Most popular questions from this chapter

A runner runs in a westerly direction on the deck of a cruise ship at \(8 \mathrm{km} / \mathrm{h}\). The cruise ship is moving north at a speed of \(35 \mathrm{km} / \mathrm{h}\). Find the velocity of the runner relative to the water.

Find the initial point of \(\mathbf{v}=(-3,1)\) if the terminal point is (5,0).

Consider the points \(A\) and \(B\) given and answer the following questions: a) Find \(|\overrightarrow{A B}|\). b) Find the components of the vector \(\mathbf{u}=\overrightarrow{A B}\) and sketch it in standard position. c) Write the vector \(\mathbf{v}=\frac{1}{|\overrightarrow{\mathrm{AB}}|} \cdot \mathbf{u}\) in component form. d) Find \(|\mathbf{v}|\). e) Sketch the vector \(\mathbf{v}\) and compare it to \(\mathbf{u}\). $$A(3,5) \text { and } B(0,5)$$

Let \(\mathbf{u}=\mathbf{i}+\mathbf{j}\) and \(\mathbf{v}=-\mathbf{i}+\mathbf{j} .\) Show that, if \(\mathbf{w}\) is any vector in the plane, then it can be written as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\). (You can generalize the result to any two non-zero, non-parallel vectors \(\mathbf{u}\) and \(\mathbf{v}\).)

Determine whether \(\mathbf{u}\) is orthogonal, parallel or neither to \(\mathbf{v}\) : $$\begin{aligned} &\mathbf{u}=\left(\begin{array}{c} -\frac{1}{2} \\ 2 \end{array}\right), \mathbf{v}=\left(\begin{array}{c} -2 \\ \frac{1}{2} \end{array}\right)\\\ & \mathbf{u}=\left(\begin{array}{l} 8 \\ 4 \end{array}\right), \mathbf{v}=\left(\begin{array}{c} 6 \\ -12 \end{array}\right)\\\ &\mathbf{u}=\left(\begin{array}{c} 2 \sqrt{3} \\ 2 \end{array}\right), \mathbf{v}=\left(\begin{array}{c} 1 \\ -\sqrt{3} \end{array}\right) \end{aligned}$$
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