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 \(\mathbf{v}=\langle 4,1\rangle\) $$-4 x=u-8 v$$

Short answer

Question: Given the equation \(-4\mathbf{x}=\mathbf{u}-8\mathbf{v}\), with \(\mathbf{u}=\langle 2, -3\rangle\) and \(\mathbf{v}=\langle 4, 1\rangle\), find the unknown vector \(\mathbf{x}\). Answer: \(\mathbf{x} = \langle 7.5, 2.75 \rangle\)

Step-by-step solution

Rewrite the equation in terms of known vectors

Given the equation, $$-4\mathbf{x}=\mathbf{u}-8\mathbf{v}$$

Substitute the known values for \(\mathbf{u}\) and \(\mathbf{v}\)

Substitute the values of \(\mathbf{u}\) and \(\mathbf{v}\): $$-4\mathbf{x}=\langle 2, -3\rangle - 8\langle 4, 1\rangle$$

Perform scalar multiplication and vector subtraction

To solve for \(\mathbf{x}\), first, we need to perform the scalar multiplication and subsequently the vector subtraction: $$-4 \mathbf{x} = \langle 2, -3\rangle - 8\langle 4, 1\rangle = \langle 2, -3\rangle - \langle 32, 8\rangle$$ $$-4 \mathbf{x} = \langle 2-32, -3-8\rangle$$

Simplify the resulting vector

Simplify the resulting vector: $$-4 \mathbf{x} = \langle -30, -11\rangle$$

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

To find \(\mathbf{x}\), divide the simplified vector by the scalar: $$\mathbf{x} = \frac{1}{-4} \langle -30, -11\rangle$$

Multiply and simplify the vector

Multiply the vector by the scalar and simplify the result: $$\mathbf{x} = \langle \frac{-30}{-4}, \frac{-11}{-4} \rangle = \langle 7.5, 2.75 \rangle$$ Hence, the unknown vector \(\mathbf{x} = \langle 7.5, 2.75 \rangle\).

Key concepts

Scalar Multiplication

When we encounter a concept such as scalar multiplication in the context of vectors, we're talking about multiplying a vector by a single number, known as a scalar. This number effectively scales the vector, either stretching or compressing it, or even reversing its direction if the scalar is negative.

For example, if we have a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \) and we multiply it by a scalar \( k \), we apply that scalar to each component of the vector to get \( k\mathbf{v} = \langle kv_1, kv_2 \rangle \). In the context of our exercise, multiplying the vector \( \mathbf{x} \) by \( -4 \) yields \( -4\mathbf{x} \), which changes both the magnitude and direction of \( \mathbf{x} \).

Vector Subtraction

Vector subtraction is another fundamental operation in vector algebra. Similar to how we perform addition, subtraction involves taking corresponding components of the vectors and subtracting them. If we have two vectors, \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) the result of their subtraction \( \mathbf{a} - \mathbf{b} \) will be \( \langle a_1-b_1, a_2-b_2 \rangle \).

In our exercise, we subtracted the vector \( 8\mathbf{v} \) from \( \mathbf{u} \), following the principle of component-wise subtraction to arrive at the new vector that is scaled by \( -4 \) in the next step.

Vector Properties

Vectors have unique properties that distinguish them from regular scalar quantities. They have both a direction and a magnitude, and their representation is often in the form of an ordered pair or tuple, such as \( \langle a, b \rangle \). The properties of vector arithmetic are similar to those of real numbers; they are commutative, associative, and distributive over addition.

Additionally, the zero vector, denoted as \( \mathbf{0} \), acts as the additive identity, and for any vector \( \mathbf{v} \) there exists a \( -\mathbf{v} \) such that \( \mathbf{v} + (-\mathbf{v}) = \mathbf{0} \). In the solution process, understanding these properties helps us to manipulate and combine vectors systematically to find the unknown vector \( \mathbf{x} \).

Unknown Vector Determination

Determining an unknown vector typically involves using known vectors and the properties of vector operations to find the vector's components. It's much like solving for an unknown in an algebraic equation but instead, we deal with each component of the vector. Steps often include rearranging the equation, scalar multiplication, addition or subtraction, and finally, isolating the unknown vector.

As observed in the exercise, to solve for \( \mathbf{x} \) we performed specific operations step by step. After simplifying the vectors through subtraction and scalar multiplication, we isolated \( \mathbf{x} \) by dividing through by the scalar. This step utilizes a key idea; vectors can be divided by scalars, reversing the effect of scalar multiplication. By correctly applying these concepts, we can solve for \( \mathbf{x} \) and find its precise components.