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Chapter 12: Problem 66

Use the properties of vectors to solve the following equations for the unknown vector \(\mathbf{x}=\langle a, b\rangle .\) Let \(\mathbf{u}=\langle 2,-3\rangle\) and \(\mathbf{v}=\langle-4,1\rangle\) $$-4 \mathbf{x}=\mathbf{u}-8 \mathbf{v}$$

Short Answer

Expert verified
Answer: The unknown vector \(\mathbf{x}\) is \(\langle 7.5, -1.25\rangle\).

Step by step solution

01

Rewrite the equation with given vectors

First, substitute the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) into the equation \(-4 \mathbf{x}=\mathbf{u}-8 \mathbf{v}\): $$-4 \mathbf{x}=\langle 2,-3\rangle - 8\langle-4,1\rangle$$
02

Calculate the value of \(\mathbf{u}-8 \mathbf{v}\)

Distribute the scalar \(-8\) to the components of vector \(\mathbf{v}\): $$-8\langle-4,1\rangle = \langle -8(-4), -8(1) \rangle = \langle 32, -8 \rangle$$ Now subtract \(\mathbf{v}\) from \(\mathbf{u}\): $$\mathbf{u}-8 \mathbf{v}= \langle 2, -3 \rangle - \langle 32, -8 \rangle = \langle 2 - 32, -3 - (-8) \rangle = \langle -30, 5 \rangle$$ Now the equation is: $$-4 \mathbf{x} = \langle -30, 5 \rangle$$
03

Divide by scalar to find \(\mathbf{x}\)

Divide both sides by \(-4\) to isolate \(\mathbf{x}\): $$\mathbf{x}=\frac{1}{-4}\langle -30, 5\rangle = \langle \frac{-30}{-4},\frac{5}{-4}\rangle$$ Simplify the components: $$\mathbf{x}=\langle 7.5, -1.25\rangle$$ Hence, the unknown vector \(\mathbf{x}\) is \(\langle 7.5, -1.25\rangle\).

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Key Concepts

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

Scalar Multiplication
Multiplying a vector by a scalar means scaling each component of the vector by the scalar. It adjusts the size but not the direction of the vector, unless the scalar is negative, which flips the direction.

In our exercise, we applied scalar multiplication when determining \( -8 \mathbf{v} \) where \( \mathbf{v} = \langle -4, 1 \rangle \). This process involved:
  • Multiplying each component by \( -8 \).
  • Resulting in the vector \( \langle 32, -8 \rangle \).
Understanding scalar multiplication is fundamental as it helps adjust vectors to desired magnitudes or directions. This principle is vital when solving vector equations.
Vector Subtraction
Vector subtraction allows us to find the difference between two vectors by subtracting corresponding components. In the problem, we used vector subtraction when finding \( \mathbf{u} - 8 \mathbf{v} \).

First, we calculated \( 8 \mathbf{v} = \langle 32, -8 \rangle \) using scalar multiplication. Then, subtracting these results from \( \mathbf{u} = \langle 2, -3 \rangle \) gave:
  • Subtract the \( x \)-components: \( 2 - 32 = -30 \)
  • Subtract the \( y \)-components: \( -3 - (-8) = 5 \)
Thus, vector subtraction can be visualized as moving along the plane from one vector to another, critically aiding in solving vector-based equations.
Vector Properties
Vectors possess several important properties that simplify algebraic manipulations, much like the properties of numbers. These include commutative, associative, and distributive properties.

Applying these properties effectively leads to streamlined solutions of vector equations. For example:
  • The distributive property, which we used in scalar multiplication, states \( a(\mathbf{b} + \mathbf{c}) = a\mathbf{b} + a\mathbf{c} \).
  • The commutative property allows the order of addition to be switched: \( \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \).
  • The associative property for addition helps group vectors conveniently: \( (\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c}) \).
Gaining a solid understanding of these properties is key to untangling more complex vector equations, helping ensure accurate solutions.

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Most popular questions from this chapter

Parabolic trajectory In Example 7 it was shown that for the parabolic trajectory \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle, \mathbf{a}=\langle 0,2\rangle\) and \(\mathbf{a}=\frac{2}{\sqrt{1+4 t^{2}}}(\mathbf{N}+2 t \mathbf{T}) .\) Show that the second expression for \(\mathbf{a}\) reduces to the first expression.

Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{4-t^{2}} \mathbf{i}+\sqrt{t} \mathbf{j}-\frac{2}{\sqrt{1+t}} \mathbf{k}$$

Prove that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$$

An object moves along a path given by \(\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle\) for \(0 \leq t \leq 2 \pi\) a. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the path is a circle (in a plane)? b. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the path is an ellipse (in a plane)?
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