Problem 25 - A Graphical Approach to Precalculus with Limits

Question

Given vectors u and v, find (a) \(2 u\) (b) \(2 u+3 v\) (c) \(v-3 u\) Do not use a calculator. $$\mathbf{u}=\langle- 1,2\rangle, \mathbf{v}=\langle 3,0\rangle$$

Step-by-Step Solution

Answer

(a) \( \langle -2, 4 \rangle \); (b) \( \langle 7, 4 \rangle \); (c) \( \langle 6, -6 \rangle \)

Step 1: Multiply vector u by 2

First, find the result of multiplying vector \( \mathbf{u} = \langle -1, 2 \rangle \) by 2. This operation involves multiplying each component of the vector by 2:\[2 \mathbf{u} = 2 \times \langle -1, 2 \rangle = \langle 2 \times -1, 2 \times 2 \rangle = \langle -2, 4 \rangle\]

Step 2: Multiply vector v by 3

Next, multiply vector \( \mathbf{v} = \langle 3, 0 \rangle \) by 3. This operation involves multiplying each component of the vector by 3:\[3 \mathbf{v} = 3 \times \langle 3, 0 \rangle = \langle 3 \times 3, 3 \times 0 \rangle = \langle 9, 0 \rangle\]

Step 3: Calculate 2u + 3v

Now calculate \( 2\mathbf{u} + 3\mathbf{v} \) by adding the results from the previous steps. You add the corresponding components of the vectors: \[2 \mathbf{u} + 3 \mathbf{v} = \langle -2, 4 \rangle + \langle 9, 0 \rangle = \langle -2 + 9, 4 + 0 \rangle = \langle 7, 4 \rangle\]

Step 4: Multiply vector u by 3

Multiply vector \( \mathbf{u} = \langle -1, 2 \rangle \) by 3. This operation involves multiplying each component of the vector by 3:\[3 \mathbf{u} = 3 \times \langle -1, 2 \rangle = \langle 3 \times -1, 3 \times 2 \rangle = \langle -3, 6 \rangle\]

Step 5: Calculate v - 3u

Finally, calculate \( \mathbf{v} - 3\mathbf{u} \) by subtracting the vector found in Step 4 from vector \( \mathbf{v} \):\[\mathbf{v} - 3\mathbf{u} = \langle 3, 0 \rangle - \langle -3, 6 \rangle = \langle 3 - (-3), 0 - 6 \rangle = \langle 6, -6 \rangle\]

Key Concepts

Vector Addition

Vector addition is like connecting arrows end-to-end in a straight path. When you're given vectors with two components, as seen with vectors \( \mathbf{u} \) and \( \mathbf{v} \), you can think of each vector component as steps in two different directions. To find the resultant vector when adding two vectors, you just need to add the corresponding components of each vector.

• Picture each component of a vector as part of a grid. The first value represents movement along the x-axis, and the second value represents movement along the y-axis.
• For example, if you add vectors \( \langle a, b \rangle \) and \( \langle c, d \rangle \), the resultant vector is \( \langle a+c, b+d \rangle \).

In the given exercise, when you add \( 2\mathbf{u} = \langle -2, 4 \rangle \) and \( 3\mathbf{v} = \langle 9, 0 \rangle \), the result is \( \langle -2+9, 4+0 \rangle = \langle 7, 4 \rangle \). You simply sum up each corresponding component to find the new position on the grid.

Scalar Multiplication

Scalar multiplication is about scaling a vector, either stretching it or shrinking it, by a certain factor. This process involves taking each component of a vector and multiplying it by the scalar value.

• If the scalar is greater than 1, the vector becomes longer. If it is less than 1, the vector becomes shorter.
• The direction of the original vector remains unchanged, but if the scalar is negative, the vector's direction reverses.

In the exercise, multiplying \( \mathbf{u} = \langle -1, 2 \rangle \) by 2 results in \( 2\mathbf{u} = \langle -2, 4 \rangle \). Similarly, multiplying \( \mathbf{v} = \langle 3, 0 \rangle \) by 3 yields \( 3\mathbf{v} = \langle 9, 0 \rangle \). Each component is adjusted depending on the scalar, stretching the vector along its path on the grid.

Vector Subtraction

Vector subtraction follows a similar principle to vector addition but in reverse. Instead of plotting your steps forward, you're moving backward by the components of one vector from another.

• Think of vector subtraction \( \mathbf{v} - \mathbf{u} \) as adding \( \mathbf{v} \) and the inverse of \( \mathbf{u} \), i.e., \( \mathbf{u} \) multiplied by \(-1\).
• The inverse operation essentially means flipping the direction of the vector you are subtracting.

For instance, to calculate \( \mathbf{v} - 3\mathbf{u} \) with given vectors, we find \( 3\mathbf{u} = \langle -3, 6 \rangle \) first, then calculate \( \mathbf{v} - 3\mathbf{u} = \langle 3, 0 \rangle - \langle -3, 6 \rangle \), resulting in \( \langle 6, -6 \rangle \). Here, each component from \( \mathbf{u} \) is subtracted from each component of \( \mathbf{v} \), thus moving us in reverse along those dimensions.