Problem 13 - Contemporary Precalculus

Question

Find \(u+v, v-u,\) and \(2 u-3 v\). $$\mathbf{u}=\langle 3,3 \sqrt{2}\rangle, \mathbf{v}=\langle 4 \sqrt{2}, 1\rangle$$

Step-by-Step Solution

Answer

Question: Given the vectors $\mathbf{u}=\langle 3,3\sqrt{2}\rangle$ and $\mathbf{v}=\langle 4\sqrt{2},1\rangle$, find the sum $\mathbf{u}+\mathbf{v}$, the difference $\mathbf{v}-\mathbf{u}$, and the linear combination $2\mathbf{u}-3\mathbf{v}$. Answer: The sum is $\mathbf{u}+\mathbf{v}=\langle 3+4\sqrt{2},3\sqrt{2}+1\rangle$, the difference is $\mathbf{v}-\mathbf{u}=\langle 4\sqrt{2}-3,1-3\sqrt{2}\rangle$, and the linear combination is $2\mathbf{u}-3\mathbf{v}=\langle 6-12\sqrt{2},6\sqrt{2}-3\rangle$.

Step 1: Vector Addition

To find the sum \(\mathbf{u} + \mathbf{v}\), we add the corresponding components of the two vectors: \((u_1 + v_1, u_2 + v_2)\). This gives us: $$ \mathbf{u}+\mathbf{v}=\langle 3+4\sqrt{2},3\sqrt{2}+1\rangle $$

Step 2: Vector Subtraction

To find the difference \(\mathbf{v} - \mathbf{u}\), we subtract the corresponding components of the two vectors: \((v_1 - u_1, v_2 - u_2)\). This gives us: $$ \mathbf{v}-\mathbf{u}=\langle 4\sqrt{2}-3,1-3\sqrt{2}\rangle $$

Step 3: Scalar Multiplication and Linear Combination

To find the linear combination \(2\mathbf{u}-3\mathbf{v}\), we first multiply each vector by their respective scalar, then subtract the second resulting vector from the first. This gives us: $$ 2\mathbf{u}-3\mathbf{v}=\langle 2(3)-3(4\sqrt{2}),2(3\sqrt{2})-3(1)\rangle=\langle 6-12\sqrt{2},6\sqrt{2}-3\rangle $$

Step 4: Final Answer

The requested results are: $$ \mathbf{u}+\mathbf{v}=\langle 3+4\sqrt{2},3\sqrt{2}+1\rangle $$ $$ \mathbf{v}-\mathbf{u}=\langle 4\sqrt{2}-3,1-3\sqrt{2}\rangle $$ $$ 2\mathbf{u}-3\mathbf{v}=\langle 6-12\sqrt{2},6\sqrt{2}-3\rangle $$

Key Concepts

Vector Addition

Vector addition is like combining two journeys to see where you end up. Imagine you have two vectors, \( \mathbf{u} = \langle 3, 3\sqrt{2} \rangle \) and \( \mathbf{v} = \langle 4\sqrt{2}, 1 \rangle \).
To add them, take the components of \( \mathbf{u} \) and \( \mathbf{v} \) and add them separately:

• First component: \( 3 + 4\sqrt{2} \)
• Second component: \( 3\sqrt{2} + 1 \)

The resulting vector is \( \mathbf{u} + \mathbf{v} = \langle 3 + 4\sqrt{2}, 3\sqrt{2} + 1 \rangle \). This new vector represents the combination of the two vectors’ directions and magnitudes.

Vector Subtraction

Vector subtraction helps you find the difference between two journeys. If \( \mathbf{v} \) is where you end, and \( \mathbf{u} \) is where you start, subtract \( \mathbf{u} \) from \( \mathbf{v} \).
Deal with each component separately as well:

• First component: \( 4\sqrt{2} - 3 \)
• Second component: \( 1 - 3\sqrt{2} \)

So, the difference is \( \mathbf{v} - \mathbf{u} = \langle 4\sqrt{2} - 3, 1 - 3\sqrt{2} \rangle \). This tells you how far and which way you need to go to get from the end of one vector to the beginning of the other.

Scalar Multiplication

Scalar multiplication involves scaling a vector by a number. It’s like stretching or shrinking a vector while keeping its direction.
For the vector \( \mathbf{u} = \langle 3, 3\sqrt{2} \rangle \), multiplying by a scalar \( 2 \):

• First component: \( 2 \times 3 = 6 \)
• Second component: \( 2 \times 3\sqrt{2} = 6\sqrt{2} \)

For \( \mathbf{v} = \langle 4\sqrt{2}, 1 \rangle \) multiplied by \( 3 \):

• First component: \( 3 \times 4\sqrt{2} = 12\sqrt{2} \)
• Second component: \( 3 \times 1 = 3 \)

Scalar multiplication adjusts the size of the vector by a consistent factor.

Linear Combination

A linear combination of vectors involves adding or subtracting scaled vectors. In this case, you form \( 2\mathbf{u} - 3\mathbf{v} \). First, find \( 2\mathbf{u} \) and \( -3\mathbf{v} \). Then add them together.
For \( 2\mathbf{u} = \langle 6, 6\sqrt{2} \rangle \) and \( -3\mathbf{v} = \langle -12\sqrt{2}, -3 \rangle \):

• First component: \( 6 - 12\sqrt{2} \)
• Second component: \( 6\sqrt{2} - 3 \)

The resulting vector is \( 2\mathbf{u} - 3\mathbf{v} = \langle 6 - 12\sqrt{2}, 6\sqrt{2} - 3 \rangle \). Linear combinations are powerful as they help combine different movements to find an overall result.