Problem 35 - Algebra and Trigonometry

Question

Find \(\operatorname{proj}_{w} \mathbf{v}\) . Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to w and \(\mathbf{v}_{2}\) is orthogonal to w. $$ \mathbf{v}=\mathbf{i}+3 \mathbf{j}, \quad \mathbf{w}=-2 \mathbf{i}+5 \mathbf{j} $$

Step-by-Step Solution

Answer

The projection of vector v onto w is \(- 8/29 \mathbf{i}+ 65/29 \mathbf{j}\). After decomposing v into two vectors: one parallel to w is \(- 8/29 \mathbf{i}+ 65/29 \mathbf{j}\) and the one orthogonal to w is \(37/29 \mathbf{i}+ 24/29 \mathbf{j}\)

Step 1: Calculating the dot product of vectors v and w

The dot product of vectors v and w, represented as \(v \cdot w\), can be calculated using the formula \(v \cdot w = \|v\|\|w\| \cos{\theta}\), where \(\|\)\ denotes the magnitude or length of the vectors. However, in this case, we can directly compute the dot product because the components of vectors v and w are given. The dot product of \(v = \mathbf{i}+3 \mathbf{j}\) and \(w = -2 \mathbf{i}+5 \mathbf{j}\) is \(v \cdot w = 1*(-2) + 3*5 = 13 \)

Step 2: Calculating the magnitude of vector w

Calculate the magnitude (also known as the length or the norm) of vector w. The magnitude is given by \(\|w\|= \sqrt{(-2)^2 + 5^2} = \sqrt{29}\)

Step 3: Finding the projection of vector v onto w

The projection of vector v onto w can be found using the formula \(\operatorname{proj}_{w} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{w}\|^2}\mathbf{w}\). Using our previous calculations, \(\operatorname{proj}_{w} \mathbf{v} = \frac{13}{29} (-2\mathbf{i} + 5\mathbf{j}) = - 8/29 \mathbf{i}+ 65/29 \mathbf{j}\)

Step 4: Decompose vector v into parallel and orthogonal vectors

As requested, vector v can be decomposed into two vectors. The vector parallel to w would be our calculated projection of v onto w, i.e. \(- 8/29 \mathbf{i}+ 65/29 \mathbf{j}\). Solve for the orthogonal vector v2 by subtracting the parallel vector from v: \(\mathbf{v}_{2} = \mathbf{v} - \mathbf{v}_{1} = (\mathbf{i}+3 \mathbf{j}) - (- 8/29 \mathbf{i}+ 65/29 \mathbf{j}) = 37/29 \mathbf{i}+ 24/29 \mathbf{j}\)

Key Concepts

Dot Product

The dot product is a way to multiply two vectors that gives us a scalar (a single number) as the result. We could think of it as a measure of how much one vector extends in the direction of another. The calculation is simple: multiply corresponding components of the two vectors and sum up those products.

For instance, given two vectors in a 2D space \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}\), their dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as \(a_1b_1 + a_2b_2\). This operation is crucial for many geometric calculations, such as finding the angle between vectors and in our exercise, for projecting one vector onto another.

Vector Decomposition

Vector decomposition involves breaking down a vector into components that are parallel and orthogonal (perpendicular) to a given vector. This process is invaluable when analyzing vectors in different directions.

To perform vector decomposition, we use the vector projection (for the parallel component) and subtract this from the original vector to find the orthogonal component. In other words, if vector \(\mathbf{v}\) is the one we're decomposing and we're using \(\mathbf{w}\) for the direction, we first find \(\mathbf{v}_{1}\) which is \(\operatorname{proj}_{\mathbf{w}} \mathbf{v}\), and then calculate \(\mathbf{v}_{2} = \mathbf{v} - \mathbf{v}_{1}\), which gives us the orthogonal vector to \(\mathbf{w}\).

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles to one another. In the Cartesian plane, this is similar to the perpendicularity of lines. Mathematically, two vectors are orthogonal if their dot product is zero. This concept is central to understanding vector projection and decomposition.

If we have vectors \(\mathbf{u}\) and \(\mathbf{v}\), and they are orthogonal, then \(\mathbf{u} \cdot \mathbf{v} = 0\). In our exercise scenario, we found a vector orthogonal to \(\mathbf{w}\) by subtracting the projection of \(\mathbf{v}\) onto \(\mathbf{w}\) from \(\mathbf{v}\) itself, ensuring the resulting vector is perpendicular to \(\mathbf{w}\), hence the term 'orthogonal vector'.

Vector Magnitude

The magnitude of a vector, often interpreted as its 'length', is a measure of how far the vector extends from its starting point. It is calculated by taking the square root of the sum of the squares of its components.

In mathematical terms, the magnitude of a vector \(\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j}\) in two dimensions is given by \(||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2}\). The vector magnitude is fundamental for normalizing vectors (making their length equal to 1), which is needed when we want to find the direction without considering the length. In our exercise, we calculated the magnitude of vector \(\mathbf{w}\) to determine the projection of \(\mathbf{v}\) onto it.