Problem 19 - Contemporary Precalculus

Question

Find \(u+v, v-u,\) and \(2 u-3 v\). $$\mathbf{u}=-\left(2 \mathbf{i}+\frac{3}{2} \mathbf{j}\right), \mathbf{v}=\frac{3}{4} \mathbf{i}$$

Step-by-Step Solution

Answer

Question: Determine the sum, difference, and linear combination of the given vectors: $$\mathbf{u}=-\left(2 \mathbf{i}+\frac{3}{2} \mathbf{j}\right)$$ and $$\mathbf{v}=\frac{3}{4} \mathbf{i}$$ Answer: The sum of the vectors is $$\mathbf{u}+\mathbf{v} = -\frac{5}{4}\mathbf{i} - \frac{3}{2}\mathbf{j}$$. The difference of the vectors is $$\mathbf{v}-\mathbf{u} = \frac{11}{4}\mathbf{i} + \frac{3}{2}\mathbf{j}$$. The linear combination of the vectors is $$2\mathbf{u}-3\mathbf{v} = -\frac{25}{4}\mathbf{i} -3\mathbf{j}$$.

Step 1: Determine the components of the given vectors

We are given the vectors: $$\mathbf{u}=-\left(2 \mathbf{i}+\frac{3}{2} \mathbf{j}\right)$$ and $$\mathbf{v}=\frac{3}{4} \mathbf{i}$$

Step 2: Sum of the vectors

To find \(u+v\), add the respective components of the vectors as follows: $$(u+v)_i=u_i+v_i=(-2)+ \frac{3}{4} = -\frac{5}{4}$$ $$(u+v)_j=u_j+v_j=-\frac{3}{2}+0=-\frac{3}{2}$$ So, $$\mathbf{u}+\mathbf{v} = -\frac{5}{4}\mathbf{i} - \frac{3}{2}\mathbf{j}$$

Step 3: Difference of the vectors

To find \(v-u\), subtract the respective components of the vectors as follows: $$(v-u)_i=v_i-u_i=\frac{3}{4}-(-2)=\frac{11}{4}$$ $$(v-u)_j=v_j-u_j=0-\left(-\frac{3}{2}\right)=\frac{3}{2}$$ So, $$\mathbf{v}-\mathbf{u} = \frac{11}{4}\mathbf{i} + \frac{3}{2}\mathbf{j}$$

Step 4: Linear combination of the vectors

To find \(2u-3v\), perform the operations as follows: $$(2u-3v)_i=2u_i-3v_i=2(-2)-3\left(\frac{3}{4}\right)=-4-\frac{9}{4}=-\frac{25}{4}$$ $$(2u-3v)_j=2u_j-3v_j=2\left(-\frac{3}{2}\right)-3(0)=-3$$ So, $$2\mathbf{u}-3\mathbf{v} = -\frac{25}{4}\mathbf{i} -3\mathbf{j}$$

Key Concepts

Understanding Vector Addition

Vector addition is a geometric operation that involves combining two or more vectors to produce a single resultant vector. It is performed by adding the corresponding components of the vectors. If you picture vectors as arrows pointing in some direction, adding vectors involves placing the tail of one arrow at the head of another.

Consider the vectors \( \mathbf{u} = -\left(2 \mathbf{i} + \frac{3}{2} \mathbf{j}\right) \) and \( \mathbf{v} = \frac{3}{4} \mathbf{i} \). The task of finding \( \mathbf{u} + \mathbf{v} \) is done by adding the i-components and j-components separately:

• The i-component is calculated as \( (-2) + \frac{3}{4} = -\frac{5}{4} \).
• The j-component is calculated as \( -\frac{3}{2} + 0 = -\frac{3}{2} \).

Thus, the resulting vector \( \mathbf{u} + \mathbf{v} \) is \( -\frac{5}{4}\mathbf{i} - \frac{3}{2}\mathbf{j} \).

Remember, vector addition is commutative, meaning \( \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} \), and it represents a physical situation where forces, velocities, or other quantities combine.

Grasping Vector Subtraction

Vector subtraction is slightly different from addition. Instead of visually placing one vector's tail at another's head, subtracting vectors involves adding a vector pointing in the opposite direction to the second vector.

To find the difference \( \mathbf{v} - \mathbf{u} \), we calculate:

• For the i-components: \( \frac{3}{4} -(-2) = \frac{11}{4} \).
• For the j-components: \( 0 - \left(-\frac{3}{2}\right) = \frac{3}{2} \).

Thus, the resulting vector \( \mathbf{v} - \mathbf{u} \) becomes \( \frac{11}{4}\mathbf{i} + \frac{3}{2}\mathbf{j} \).

Subtraction can be visualized by thinking of "flipping" one vector's direction and then adding it to another vector. This operation illustrates how differences in direction or magnitude between forces or velocities are resolved.

Defining Linear Combination of Vectors

Linear combination refers to forming a new vector by multiplying given vectors by constants (scalars) and then adding the resulting vectors together. This is a fundamental operation within vector spaces, allowing one to scale and combine vectors.

For the expression \( 2\mathbf{u} - 3\mathbf{v} \), we scale each vector by the respective constant and then perform vector subtraction:

• Calculate \( 2 \times (-2) = -4 \) for the i-component of \( \mathbf{u} \) and \(-3 \times \frac{3}{4} = -\frac{9}{4} \) for \( \mathbf{v} \), resulting in: \(-4 - \frac{9}{4} = -\frac{25}{4} \).
• For the j-component of \( \mathbf{u} \), \( 2 \times -\frac{3}{2} = -3 \), since the j-component of \( \mathbf{v} \) is zero: \(-3 - 0 = -3 \).

Consequently, the vector \( 2\mathbf{u} - 3\mathbf{v} \) simplifies to \( -\frac{25}{4}\mathbf{i} - 3\mathbf{j} \).

Understanding linear combinations is key when dealing with problems involving span, basis, and linear transformations in vector spaces. This operation not only extends basic vector arithmetic but also underpins more advanced mathematical concepts.