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Vector addition is a fundamental operation in vector algebra. It involves adding two or more vectors together to produce a resultant vector. Each vector is characterized by its components. To add vectors, you simply add the corresponding components from each vector.
For example, if you have two vectors, \( \mathbf{u} = \langle a_1, b_1 \rangle \) and \( \mathbf{v} = \langle a_2, b_2 \rangle \), vector addition is done component-wise:

• Add the first components: \( a_1 + a_2 \)
• Add the second components: \( b_1 + b_2 \)

So, the sum of the vectors is: \( \mathbf{u} + \mathbf{v} = \langle a_1 + a_2, b_1 + b_2 \rangle \).
In our given exercise, \( \mathbf{u} = \langle \frac{2}{3}, 4 \rangle \) and \( \mathbf{v} = \langle -7, \frac{19}{3} \rangle \). Using the rule, to find \( \mathbf{u} + \mathbf{v} \), you calculate:

• First component: \( \frac{2}{3} + (-7) = -\frac{19}{3} \)
• Second component: \( 4 + \frac{19}{3} = \frac{31}{3} \)

Thus, \( \mathbf{u} + \mathbf{v} = \langle -\frac{19}{3}, \frac{31}{3} \rangle \). This operation helps in solving many problems in physics and engineering where forces, velocities, or other vector quantities need to be combined.

Vector subtraction is a similar process to vector addition, but instead of adding the components, you subtract them. This operation gives us the resultant vector pointing from one vector to another. Just like addition, vector subtraction works component by component.
Take two vectors, \( \mathbf{u} = \langle a_1, b_1 \rangle \) and \( \mathbf{v} = \langle a_2, b_2 \rangle \). To subtract vector \( \mathbf{u} \) from vector \( \mathbf{v} \), perform:

• Subtract the first components: \( a_2 - a_1 \)
• Subtract the second components: \( b_2 - b_1 \)

The resultant vector is \( \mathbf{v} - \mathbf{u} = \langle a_2 - a_1, b_2 - b_1 \rangle \).
For our exercise, the given vectors \( \mathbf{u} = \langle \frac{2}{3}, 4 \rangle \) and \( \mathbf{v} = \langle -7, \frac{19}{3} \rangle \) are used as:

• First component: \( -7 - \frac{2}{3} = -\frac{23}{3} \)
• Second component: \( \frac{19}{3} - 4 = \frac{7}{3} \)

Resulting in \( \mathbf{v} - \mathbf{u} = \langle -\frac{23}{3}, \frac{7}{3} \rangle \). Understanding vector subtraction is crucial in navigation, physics, and any field requiring direction and magnitude adjustments.

Scalar multiplication involves multiplying a vector by a scalar (a real number). This operation scales the vector without changing its direction, unless the scalar is negative, which also reverses its direction. Each component of the vector is multiplied by the scalar.
Consider a vector \( \mathbf{u} = \langle a, b \rangle \) and a scalar \( k \). Scalar multiplication is performed as follows:

• Multiply the first component by the scalar: \( k \cdot a \)
• Multiply the second component by the scalar: \( k \cdot b \)

Resulting in: \( k \mathbf{u} = \langle k \cdot a, k \cdot b \rangle \).
In the given exercise, you need to compute \( 2\mathbf{u} - 3\mathbf{v} \). This involves two separate scalar multiplications followed by vector subtraction:

• First, multiply \( \mathbf{u} = \langle \frac{2}{3}, 4 \rangle \) by 2: \( 2\mathbf{u} = \langle \frac{4}{3}, 8 \rangle \)
• Next, multiply \( \mathbf{v} = \langle -7, \frac{19}{3} \rangle \) by 3: \( 3\mathbf{v} = \langle -21, \frac{57}{3} \rangle \)
• Now, perform vector subtraction: \( 2\mathbf{u} - 3\mathbf{v} = \langle \frac{4}{3} + 21, 8 - \frac{57}{3} \rangle = \langle \frac{31}{3}, -\frac{25}{3} \rangle \)

Scalar multiplication is used widely in computer graphics, physics, and engineering to adjust vector quantities according to size or magnitude.