Question

Solve the following pairs of equations for the vectors \(\mathbf{u}\) and \(\mathbf{v} .\) Assume \(\mathbf{i}=\langle 1,0\rangle\) and \(\mathbf{j}=\langle 0,1\rangle\) $$2 \mathbf{u}=\mathbf{i}, \mathbf{u}-4 \mathbf{v}=\mathbf{j}$$

Short answer

Question: Find the vectors 𝑢 and 𝑣 that satisfy the given vector equations: 2𝑢=𝑖 and 𝑢−4𝑣=𝑗. Answer: The vectors 𝑢 and 𝑣 that satisfy the given equations are: 𝑢 = ⟨(1/2), 0⟩ and 𝑣 = ⟨(1/8), -(1/4)⟩.

Step-by-step solution

Write the vector equations as component equations

Given the vector equations: $$2 \mathbf{u} = \mathbf{i},\qquad \mathbf{u} - 4 \mathbf{v} = \mathbf{j}$$ We can write the components of the vectors \(\mathbf{u}=\langle u_1,u_2\rangle\) and \(\mathbf{v}=\langle v_1,v_2\rangle\). Then the given equations become: $$2 \langle u_1,u_2\rangle = \langle 1,0\rangle ,\qquad \langle u_1,u_2\rangle - 4 \langle v_1,v_2\rangle = \langle 0,1\rangle $$ Now, we can write these equations as component-wise equations: $$ 2u_1 = 1,\quad 2u_2 = 0,\quad u_1 - 4v_1 = 0,\quad u_2 - 4v_2 = 1 $$

Solve the system of equations

Now, let's solve this system of linear equations for \(u_1, u_2, v_1, v_2\). From the first equation: \(2u_1=1\), we get \(u_1 = \frac{1}{2}\). From the second equation: \(2u_2=0\), we get \(u_2 = 0\). Next, substitute the value of \(u_1\) into the third equation: \(\frac{1}{2} - 4v_1 = 0\), which gives \(v_1 = \frac{1}{8}\). Finally, substitute the value of \(u_2\) into the fourth equation: \(0 - 4v_2 = 1\), which gives \(v_2 = -\frac{1}{4}\).

Write the solution in terms of vectors

Now that we have the components of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), we can rewrite them as vectors: $$ \mathbf{u} = \left\langle \frac{1}{2}, 0 \right\rangle,\quad \mathbf{v} = \left\langle \frac{1}{8}, -\frac{1}{4} \right\rangle $$ So the solution to the given vector equations is: $$\mathbf{u} = \left\langle \frac{1}{2},0 \right\rangle,\qquad \mathbf{v} = \left\langle \frac{1}{8}, -\frac{1}{4} \right\rangle $$

Key concepts

Vector Components

A vector can be understood as an arrow pointing from one position to another in space, which can be split into parts called components. For instance, in two-dimensional space, any vector can be represented as a combination of two basis vectors, usually represented by \(\mathbf{i}\) and \(\mathbf{j}\).
These correspond to movement along the x-axis and y-axis respectively.
For a vector \(\mathbf{u} = \langle u_1, u_2 \rangle\), the components \(u_1\) and \(u_2\) show how much the vector stretches in the directions of \(\mathbf{i}\) and \(\mathbf{j}\).
This allows us to write more complex expressions for vectors in terms of these individual parts, making equations involving vectors easier to handle.

• Vector components are crucial for simplifying problems into more manageable parts.
• The components of the vector are scalars that measure how far and in what direction a point moves from the origin.
• Using components, vector problems can be transformed into algebraic problems.

System of Linear Equations

A system of linear equations consists of two or more linear equations with multiple variables. The goal is to find the value of the variables that satisfy all equations simultaneously.
In the vector context, we use these systems to find specific components of vectors. For example, if \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\), solving for \(u_1, u_2, v_1,\) and \(v_2\) can tell us the exact direction and length of each vector involved.

• Solving a system often involves substitution or elimination, reducing multiple equations to one or two equations with fewer variables.
• An equation like \(2u_1 = 1\) directly gives the component, helping to build towards solving the full system.
• Careful manipulation of these equations can lead to finding accurate vector components, which provide full solutions.

Solution of Vectors

Once the calculations are complete, the solution of vectors involves expressing the vectors in terms of their components.
In our example, after solving the equations, the results are \(\mathbf{u} = \langle \frac{1}{2}, 0 \rangle\) and \(\mathbf{v} = \langle \frac{1}{8}, -\frac{1}{4} \rangle\).
This means that vector \(\mathbf{u}\) moves half a unit along the x-axis with no movement along the y-axis, while vector \(\mathbf{v}\) moves an eighth of a unit along the x-axis and a quarter of a unit in the negative direction along the y-axis.

• The results give the directional and magnitude components of each vector, defining their exact position in space.
• By solving the equations, we turn abstract vector equations into concrete, understandable numerical vectors.
• This solution helps in visualising the vectors' effects and how they contribute to the system they are part of.