Problem 62 - Precalculus with Limits

Question

In Exercises 59-62, find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\). Then write \(\mathbf{u}\) as the sum of two orthogonal vectors, one of which is proj\(_{\mathbf{v}} \mathbf{u}\). \(\mathbf{u} = \langle -3, -2 \rangle\) \(\mathbf{v} = \langle -4, -1 \rangle\)

Step-by-Step Solution

Answer

The projection of \mathbf{u} onto \mathbf{v} is \langle -\frac{56}{17}, -\frac{14}{17} \rangle. The decomposition of \mathbf{u} into two orthogonal vectors yields \mathbf{u} = \langle -\frac{56}{17}, -\frac{14}{17} \rangle + \langle \frac{1}{17}, -\frac{20}{17} \rangle.

Step 1: Calculate the projection of \mathbf{u} onto \mathbf{v}

The formula for the projection of \mathbf{u} onto \mathbf{v} is given as proj\(_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^2}\) \(\mathbf{v}\). Substituting the given vectors \mathbf{u} = \langle -3, -2 \rangle and \(\mathbf{v} = \langle -4, -1 \rangle\) into the formula gives proj\(_{\mathbf{v}} \mathbf{u} = \frac{(-3)(-4) + (-2)(-1)}{(-4)^2 + (-1)^2} \langle -4, -1 \rangle\), which simplifies to proj\(_{\mathbf{v}} \mathbf{u} = \frac{14}{17} \langle -4, -1 \rangle\), or proj\(_{\mathbf{v}} \mathbf{u} = \langle -\frac{56}{17}, -\frac{14}{17} \rangle\).

Step 2: Write \mathbf{u} as the sum of two orthogonal vectors

Since the projection of \mathbf{u} onto \mathbf{v} is one of the vectors, the other vector is found by subtracting this projection from \mathbf{u}. Therefore, the other vector, say \mathbf{w}, is given as \mathbf{w} = \mathbf{u} - proj\(_{\mathbf{v}} \mathbf{u}\). Substituting the values \mathbf{u} = \langle -3, -2 \rangle and proj\(_{\mathbf{v}} \mathbf{u} = \langle -\frac{56}{17}, -\frac{14}{17} \rangle\) gives \mathbf{w} = \langle -3, -2 \rangle - \langle -\frac{56}{17}, -\frac{14}{17} \rangle\), or \mathbf{w} = \langle -3 + \frac{56}{17}, -2 + \frac{14}{17} \rangle = \langle \frac{1}{17}, -\frac{20}{17} \rangle\). Therefore, the expression for \mathbf{u} as the sum of two orthogonal vectors is given as \mathbf{u} = proj\(_{\mathbf{v}} \mathbf{u} + \mathbf{w} = \langle -\frac{56}{17}, -\frac{14}{17} \rangle + \langle \frac{1}{17}, -\frac{20}{17} \rangle\).

Key Concepts

Orthogonal Vectors

To understand vector projection, it's essential to grasp the concept of orthogonal vectors. Two vectors are said to be orthogonal when they are perpendicular to each other, which means their direction forms a right angle. In terms of their dot product, which we'll discuss shortly, orthogonal vectors have a dot product of zero. This is because the dot product involves both the magnitudes of the vectors and the cosine of the angle between them. If the vectors are at a right angle, the cosine of 90 degrees (or \(\pi/2\) radians) is zero.

Let's illustrate this with a couple of two-dimensional vectors \(\mathbf{x}\) and \(\mathbf{y}\). If \(\mathbf{x} \cdot \mathbf{y} = 0\), then we can say that \(\mathbf{x}\) and \(\mathbf{y}\) are orthogonal. In the context of our exercise, after computing the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), the second vector (\(\mathbf{w}\)) is orthogonal to \(\mathbf{v}\), demonstrated by the fact that \(\mathbf{v}\) and \(\mathbf{w}\) do not influence each other's direction.

Dot Product

The dot product is a crucial operation in vector algebra, also known as the scalar product since it results in a scalar quantity. It is calculated by multiplying corresponding components of two vectors and then summing those products. Mathematically, the dot product of vectors \( \mathbf{a}\) and \(\mathbf{b}\) is denoted as \( \mathbf{a} \cdot \mathbf{b}\).

For two-dimensional vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), the formula is \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\). The dot product tells us about the relationship between the directions of the vectors; it is positive when the angle is acute, negative when the angle is obtuse, and zero when the vectors are orthogonal. In our exercise, the dot product was used to calculate the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), showing how \(\mathbf{u}\) relates directionally to \(\mathbf{v}\).

Vector Operations

Vector operations include addition, subtraction, and multiplication by a scalar. These operations are fundamental when it comes to manipulating vectors in various applications such as physics, engineering, and computer graphics.

For vector addition and subtraction, corresponding components are simply added or subtracted. Taking two vectors \(\mathbf{p} = \langle p_1, p_2 \rangle\) and \(\mathbf{q} = \langle q_1, q_2 \rangle\), their sum is \(\mathbf{p} + \mathbf{q} = \langle p_1 + q_1, p_2 + q_2 \rangle\), and their difference is \(\mathbf{p} - \mathbf{q} = \langle p_1 - q_1, p_2 - q_2 \rangle\). Multiplication by a scalar involves multiplying each component of the vector by the scalar. If \(\lambda\) is a scalar, then the product with \(\mathbf{q}\) is \(\lambda \mathbf{q} = \langle \lambda q_1, \lambda q_2 \rangle\).

In the textbook solution, we subtracted the projection vector from \(\mathbf{u}\) to find the orthogonal vector \(\mathbf{w}\), an operation that rearranges the components of \(\mathbf{u}\) such that the resultant vector is perpendicular to \(\mathbf{v}\). This subtraction is a key example of vector operations in action.