Problem 27 - Algebra and Trigonometry

Question

23-28 (a) Calculate proj, \(\mathbf{u} .\) (b) Resolve \(\mathbf{u}\) into \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v}\). $$\mathbf{u}=\langle 2,9\rangle, \quad \mathbf{v}=\langle- 3,4\rangle$$

Step-by-Step Solution

Answer

proj \(\mathbf{u} = \langle -\frac{18}{5}, \frac{24}{5} \rangle\); \(\mathbf{u}_1 = \langle -\frac{18}{5}, \frac{24}{5} \rangle\), \(\mathbf{u}_2 = \langle \frac{28}{5}, \frac{21}{5} \rangle\).

Step 1: Calculate the Dot Product

The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2\). Substitute \(\mathbf{u} = \langle 2, 9 \rangle\) and \(\mathbf{v} = \langle -3, 4 \rangle\) to calculate the dot product:\[\mathbf{u} \cdot \mathbf{v} = (2)(-3) + (9)(4) = -6 + 36 = 30.\]

Step 2: Calculate Magnitude of Vector v

The magnitude of a vector \(\mathbf{v} = \langle v_1, v_2 \rangle\) is given by \(||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2}\). Substitute \(\mathbf{v} = \langle -3, 4 \rangle\):\[||\mathbf{v}||^2 = (-3)^2 + 4^2 = 9 + 16 = 25.\]Hence, \(||\mathbf{v}|| = \sqrt{25} = 5\).

Step 3: Calculate Projection of u onto v

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is calculated using the formula:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \mathbf{v}.\]Substitute the given values:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{30}{25}\langle -3, 4 \rangle = \frac{6}{5}\langle -3, 4 \rangle = \langle -\frac{18}{5}, \frac{24}{5} \rangle.\]

Step 4: Resolve u into Parallel and Orthogonal Components

The parallel component \(\mathbf{u}_1\) is the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), which we have calculated as:\[\mathbf{u}_1 = \langle -\frac{18}{5}, \frac{24}{5} \rangle.\]The orthogonal component \(\mathbf{u}_2\) can be found using the equation:\[\mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1.\]Substitute the known values:\[\mathbf{u}_2 = \langle 2, 9 \rangle - \langle -\frac{18}{5}, \frac{24}{5} \rangle = \langle 2 + \frac{18}{5}, 9 - \frac{24}{5} \rangle.\]Convert to a common fraction:\[\mathbf{u}_2 = \langle \frac{10}{5} + \frac{18}{5}, \frac{45}{5} - \frac{24}{5} \rangle = \langle \frac{28}{5}, \frac{21}{5} \rangle.\]

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It combines two vectors into a single scalar value. This operation is useful for determining how much one vector extends in the direction of another.

To find the dot product of two vectors, \mathbf{u} = \langle u_1, u_2 \rangle\ and \mathbf{v} = \langle v_1, v_2 \rangle\, you calculate: \[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2. \]

• In our example: \mathbf{u} = \langle 2, 9 \rangle\ and \mathbf{v} = \langle -3, 4 \rangle\, lead to \( 2(-3) + 9(4) = -6 + 36 = 30 \).
• The result is 30, which is a scalar.

The dot product is especially informative as it also helps to find angles between vectors. If the dot product is zero, the vectors are orthogonal (perpendicular) to each other.

Magnitude of a Vector

Understanding a vector's magnitude is crucial. This tells us how long the vector is. Knowing the length of a vector can be particularly useful in mechanics and real-world applications involving direction and speed.

The magnitude of a vector \mathbf{v} = \langle v_1, v_2 \rangle\ can be calculated using the Pythagorean theorem: \[ ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2}. \]

• For vector \mathbf{v} = \langle -3, 4 \rangle\, calculate: \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.
• This means vector \mathbf{v} has a magnitude of 5 units.

Magnitude is always non-negative and it gives us the scale of the vector.

Orthogonal Components

Resolving vectors into orthogonal components is breaking the vector into parts that are perpendicular to each other. This is useful in understanding how a vector behaves in different directions.

When you resolve \mathbf{u} into components parallel and orthogonal to \mathbf{v}, you are technically decomposing \mathbf{u} into:

• One part, \mathbf{u}_1, that runs in the same direction as \mathbf{v} (this is known as the projection).
• Another part, \mathbf{u}_2, that is perpendicular to \mathbf{v}.

The orthogonal component can be calculated by subtracting the parallel component from the original vector: \[ \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \]
In our case, \mathbf{u}_1 = \langle -\frac{18}{5}, \frac{24}{5} \rangle\ and the orthogonal component then becomes \mathbf{u}_2 = \langle \frac{28}{5}, \frac{21}{5} \rangle\.

These orthogonal components are vital in physics and engineering for analyzing forces and vectors in different directions without interference from each other.