Problem 19 - Algebra and Trigonometry

Question

Find \(2 \mathbf{u},-3 \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(3 \mathbf{u}-4 \mathbf{v}\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v} .\) $$ \mathbf{u}=\langle 0,-1\rangle, \quad \mathbf{v}=\langle- 2,0\rangle $$

Step-by-Step Solution

Answer

\(2\mathbf{u} = \langle 0, -2 \rangle\), \(-3\mathbf{v} = \langle 6, 0 \rangle\), \(\mathbf{u}+\mathbf{v} = \langle -2, -1 \rangle\), \(3\mathbf{u}-4\mathbf{v} = \langle 8, -3 \rangle\).

Step 1: Calculate 2u

To find \(2 \mathbf{u}\), multiply each component of \(\mathbf{u} = \langle 0, -1 \rangle\) by 2. Thus, \(2 \mathbf{u} = 2 \times \langle 0, -1 \rangle = \langle 2 \times 0, 2 \times (-1) \rangle = \langle 0, -2 \rangle\).

Step 2: Calculate -3v

To find \(-3 \mathbf{v}\), multiply each component of \(\mathbf{v} = \langle -2, 0 \rangle\) by -3. Thus, \(-3 \mathbf{v} = -3 \times \langle -2, 0 \rangle = \langle -3 \times (-2), -3 \times 0 \rangle = \langle 6, 0 \rangle\).

Step 3: Calculate u+v

To find \(\mathbf{u} + \mathbf{v}\), add the corresponding components of \(\mathbf{u} = \langle 0, -1 \rangle\) and \(\mathbf{v} = \langle -2, 0 \rangle\). Thus, \(\mathbf{u} + \mathbf{v} = \langle 0 + (-2), -1 + 0 \rangle = \langle -2, -1 \rangle\).

Step 4: Calculate 3u-4v

To find \(3 \mathbf{u} - 4 \mathbf{v}\), first calculate \(3 \mathbf{u}\) and \(4 \mathbf{v}\) separately, then subtract the corresponding components. First, \(3 \mathbf{u} = 3 \times \langle 0, -1 \rangle = \langle 0, -3 \rangle\). Next, \(4 \mathbf{v} = 4 \times \langle -2, 0 \rangle = \langle -8, 0 \rangle\). Now, subtract \(4 \mathbf{v}\) from \(3 \mathbf{u}\): \(3 \mathbf{u} - 4 \mathbf{v} = \langle 0 - (-8), -3 - 0 \rangle = \langle 8, -3 \rangle\).

Key Concepts

vector addition

Vector addition is a fundamental operation in vector mathematics, which allows us to combine two vectors into a single resultant vector. When adding vectors, the idea is to add the corresponding components together to find the resultant vector.

In the given exercise, we have vectors \(\mathbf{u} = \langle 0, -1 \rangle\) and \(\mathbf{v} = \langle -2, 0 \rangle\). To add these vectors, you perform the following steps:

• Add the first components of each vector: \(0 + (-2) = -2\)
• Add the second components of each vector: \(-1 + 0 = -1\)

Thus, the resultant vector from adding \(\mathbf{u}\) and \(\mathbf{v}\) is \(\langle -2, -1 \rangle\).
This means that if you visualize these vectors, \(\langle -2, -1 \rangle\) would be the endpoint of the path if you started from one vector's tail and finished by adding the other's tip.

scalar multiplication

Scalar multiplication is an operation where a vector is multiplied by a scalar (a real number). This operation changes the magnitude of the vector but not its direction, assuming the scalar is positive. If the scalar is negative, the direction of the resultant vector will also change to the opposite.

Let's see how it works in the context of the given exercise:

• For \(2 \mathbf{u}\), you multiply each component of \(\mathbf{u} = \langle 0, -1 \rangle\) by 2. This results in \(\langle 0, -2 \rangle\).
• Similarly, for \(-3 \mathbf{v}\), you multiply \(\mathbf{v} = \langle -2, 0 \rangle\) by -3, resulting in \(\langle 6, 0 \rangle\).

The resultant vectors \(\langle 0, -2 \rangle\) and \(\langle 6, 0 \rangle\) are the scaled versions of their respective original vectors. Here, you can see how the signs of the scalars affect the orientation of the vector.

vector subtraction

Vector subtraction allows us to find the difference between two vectors. This operation is similar to vector addition but involves subtracting the corresponding components of the vectors. Vector subtraction can be visualized as adding the negative of one vector to another.

In the problem given, consider the expression \(3 \mathbf{u} - 4 \mathbf{v}\):

• First, find \(3 \mathbf{u}\) by multiplying \(\mathbf{u} = \langle 0, -1 \rangle\) by 3, getting \(\langle 0, -3 \rangle\).
• Next, find \(-4 \mathbf{v}\) by multiplying \(\mathbf{v} = \langle -2, 0 \rangle\) by -4, giving \(\langle -8, 0 \rangle\).
• Finally, subtract these resulting vectors: \(\langle 0 - (-8), -3 - 0 \rangle = \langle 8, -3 \rangle\).

This result implies that, effectively, \(3 \mathbf{u} - 4 \mathbf{v}\) is the vector resulting from moving along \(3 \mathbf{u}\) and then reversing a path defined by \(4 \mathbf{v}\). Subtraction, therefore, helps us to adjust one vector by the extent (magnitude) of another.