Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 23-28, find each of the given projections if \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}, \mathbf{v}=2 \mathbf{i}-\mathbf{j}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}\). $$ \operatorname{proj}_{\mathbf{u}} \mathbf{v} $$

Step-by-Step Solution

Answer

The projection of \( \mathbf{v} \) onto \( \mathbf{u} \) is \( \mathbf{0} \).

Step 1: Calculate Dot Product of u and v

Find the dot product of the vectors \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} \) and \( \mathbf{v} = 2 \mathbf{i} - \mathbf{j} \). Use the formula \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). Here, \( u_1 = 1, u_2 = 2, v_1 = 2, v_2 = -1 \). Thus, \( \mathbf{u} \cdot \mathbf{v} = 1 \times 2 + 2 \times (-1) = 2 - 2 = 0 \).

Step 2: Calculate Magnitude of u Squared

Calculate the magnitude of \( \mathbf{u} \) squared, which is used in the projection formula. The formula is \( \| \mathbf{u} \|^2 = u_1^2 + u_2^2 \). So, \( \| \mathbf{u} \|^2 = 1^2 + 2^2 = 1 + 4 = 5 \).

Step 3: Apply Projection Formula

Use the projection formula for the vector \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \|^2} \mathbf{u} \). Since \( \mathbf{u} \cdot \mathbf{v} = 0 \) and \( \| \mathbf{u} \|^2 = 5 \), the projection is \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{0}{5} (\mathbf{i} + 2 \mathbf{j}) = 0 \mathbf{i} + 0 \mathbf{j} = \mathbf{0} \).

Key Concepts

Dot Product

In vector mathematics, the dot product (also known as the scalar product) is a fundamental operation that combines two vectors to produce a scalar, which is a single number. This operation helps to determine how much one vector goes in the same direction as the other. To compute the dot product of two vectors \( \mathbf{u} = u_1 \mathbf{i} + u_2 \mathbf{j} \) and \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} \), we use the formula: - \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \) For the vectors in the exercise, we have: - \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} \) with components \( (1, 2) \) - \( \mathbf{v} = 2 \mathbf{i} - \mathbf{j} \) with components \( (2, -1) \) Plugging these values into the formula gives \( \mathbf{u} \cdot \mathbf{v} = 1 \times 2 + 2 \times (-1) = 2 - 2 = 0 \). The resulting dot product is zero, indicating that vectors \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal, meaning they are perpendicular to each other.

Magnitude of a Vector

The magnitude (or length) of a vector is a measure of its 'size' and is similar to finding the length of a line segment. It's a scalar quantity and not a vector, representing the distance from the origin of the vector in its direction. For a vector \( \mathbf{u} = u_1 \mathbf{i} + u_2 \mathbf{j} \), the magnitude is found using the formula: - \( \| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2} \) In the exercise, the vector \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} \) has components \( (1, 2) \). First, compute the magnitude squared as: - \( \| \mathbf{u} \|^2 = 1^2 + 2^2 = 1 + 4 = 5 \) This squared form is particularly useful for calculations involving projections, as it avoids unnecessary square root operations. Knowing the magnitude is crucial when determining directions and normalizing vectors, which converts a vector to a unit vector.

Projection Formula

Vector projection is a way of representing one vector along the line of another vector and determining its contribution in that direction. It uses the concept of the dot product and is often applied in physics and engineering to find components of forces or velocities. The formula for projecting vector \( \mathbf{v} \) onto vector \( \mathbf{u} \) is given by: - \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \|^2} \mathbf{u} \) In our original problem, we first computed \( \mathbf{u} \cdot \mathbf{v} = 0 \) and \( \| \mathbf{u} \|^2 = 5 \). Substituting into the projection formula: - \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{0}{5} (\mathbf{i} + 2 \mathbf{j}) = \mathbf{0} \) This result shows that \( \mathbf{v} \) has no component along the direction of \( \mathbf{u} \), which aligns with our earlier finding of the vectors being orthogonal. Learning to use the projection formula helps understand how vectors can affect or influence each other in space, simplifying many problems in vector calculus.