Question

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

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

Step-by-step solution

Rewrite the equation with given vectors

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

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: $$-4 \mathbf{x} = \langle -30, 5 \rangle$$

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 \(\langle 7.5, -1.25\rangle\).

Key concepts

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.