Problem 31 - Essential Calculus Early Transcendentals

Question

\(29-32\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\langle 3,6,-2\rangle, \quad \mathbf{b}=\langle 1,2,3\rangle$$

Step-by-Step Solution

Answer

Scalar projection is \( \frac{9}{7} \), vector projection is \( \left\langle \frac{27}{49}, \frac{54}{49}, \frac{-18}{49} \right\rangle \).

Step 1: Find the Dot Product

The first step is to find the dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \). The dot product is calculated using the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Substituting the given vectors, we have:\[\mathbf{a} \cdot \mathbf{b} = 3 \times 1 + 6 \times 2 + (-2) \times 3 = 3 + 12 - 6 = 9\]

Step 2: Find the Magnitude of Vector a

Next, we calculate the magnitude of vector \( \mathbf{a} \) using the formula \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \). Substitute the components of \( \mathbf{a} \):\[\|\mathbf{a}\| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\]

Step 3: Calculate Scalar Projection of b onto a

The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by:\[ \text{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|}\]Substitute the values calculated in Steps 1 and 2:\[\text{proj}_{\mathbf{a}} \mathbf{b} = \frac{9}{7}\]

Step 4: Calculate Vector Projection of b onto a

The vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is calculated using the formula:\[ \mathbf{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2} \right) \mathbf{a}\]Using the result from Step 1 and the magnitude squared from Step 2, \( \|\mathbf{a}\|^2 = 49 \), we find:\[\mathbf{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{9}{49} \right) \langle 3,6,-2 \rangle = \langle \frac{27}{49}, \frac{54}{49}, \frac{-18}{49} \rangle\]

Key Concepts

Dot Product

The dot product is a fundamental operation for vectors, especially in 3D space. It involves multiplying corresponding components of two vectors and summing the results. Given vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the formula is:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]The dot product gives you a scalar, not a vector. It's worth noting that when the dot product is zero, the vectors are perpendicular. In this exercise, the dot product is calculated as:- \(3 \times 1 + 6 \times 2 + (-2) \times 3 = 9\)This result gives us a sense of how much \( \mathbf{b} \) aligns with \( \mathbf{a} \). A positive value indicates that they point in a generally similar direction.

Magnitude of a Vector

The magnitude, or length, of a vector is like measuring the length of a line segment between its tail and head when visualized geometrically. For vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), it is calculated using the formula:\[\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]This is derived from the Pythagorean theorem and ensures a non-negative result, reflecting the idea of length which cannot be negative.In this problem:- \( \|\mathbf{a}\| = \sqrt{3^2 + 6^2 + (-2)^2} = 7\)A larger magnitude means a longer vector, whereas a smaller magnitude indicates a shorter vector. This concept is essential for understanding how much of one vector gets projected onto another when finding projections.

Scalar Projection

Scalar projection helps to determine the extent to which one vector points in the direction of another. It translates a vector into a simple scalar value that captures the component of one vector that "falls onto" another. The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is calculated using:\[ \text{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|} \]Using this formula:- With a dot product of 9 and the magnitude of vector \( \mathbf{a} \) as 7, the projection value is \( \frac{9}{7} \).This value tells us the "shadow" of \( \mathbf{b} \) onto \( \mathbf{a} \), i.e., how much of \( \mathbf{b} \) aligns with the direction of \( \mathbf{a} \). The scalar projection is important for decomposing vectors and understanding their behaviour relative to other directions.