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Chapter 12: Problem 38

\(35-40\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\langle- 2,3,-6\rangle, \quad \mathbf{b}=\langle 5,-1,4\rangle$$

Short Answer

Expert verified
Scalar projection is \(-\frac{37}{7}\) and vector projection is \(\langle \frac{74}{49}, -\frac{111}{49}, \frac{222}{49} \rangle\).

Step by step solution

01

Calculate the Dot Product

To find scalar and vector projections, first calculate the dot product of the vectors \( \mathbf{a} = \langle -2, 3, -6 \rangle \) and \( \mathbf{b} = \langle 5, -1, 4 \rangle \). The dot product is given by: \( \mathbf{a} \cdot \mathbf{b} = (-2)(5) + (3)(-1) + (-6)(4) = -10 - 3 - 24 = -37 \).
02

Calculate the Magnitude of Vector a

The magnitude of \( \mathbf{a} \) is needed for both projections. Compute it using the formula \( \|\mathbf{a}\| = \sqrt{(-2)^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).
03

Find the Scalar Projection of b onto a

The scalar projection (or component) of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by \( \mathrm{comp}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|} = \frac{-37}{7} = -\frac{37}{7} \).
04

Find the Vector Projection of b onto a

The vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by: \( \mathrm{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2} \mathbf{a} \). Compute: \( \frac{-37}{49} \times \langle -2, 3, -6 \rangle = \langle \frac{74}{49}, -\frac{111}{49}, \frac{222}{49} \rangle \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a fundamental operation in vector calculus that combines two vectors to produce a scalar. It's akin to multiplying the corresponding components of two vectors and then adding up the results. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is calculated as: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
  • This operation is handy for determining the angle between two vectors. If the dot product is zero, the vectors are perpendicular.
  • In this exercise, the dot product \( \mathbf{a} \cdot \mathbf{b} = -37 \), results from \(-10 - 3 - 24 = -37\).
The sign of the dot product can tell you the relative direction of the vectors. A negative result, like here, indicates they point in more opposite than same directions.
Vector Magnitude
The magnitude of a vector provides a measure of its length. It's a crucial part of vector calculations and is used, for instance, in finding projections. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude \( \|\mathbf{a}\| \) is calculated as: \[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
  • This formula is derived from the Pythagorean theorem in three-dimensional space.
  • In the problem, \( \|\mathbf{a}\| = 7 \), calculated by \( \sqrt{4 + 9 + 36} = \sqrt{49} \).
Knowing the magnitude is important as it normalizes the vector, allowing for the calculation of direction without altering its orientation.
Scalar Projection
The scalar projection of one vector onto another gives the magnitude of the projection of one vector onto the other. In simpler terms, think of it as "casting a shadow" of a vector in the direction of another. For the scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \), use the formula: \[ \mathrm{comp}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|} \]
  • This formula represents how much of \( \mathbf{b} \) extends in the direction of \( \mathbf{a} \).
  • In our case, the scalar projection is \(-\frac{37}{7} \), indicating both the length and direction of the projection.
Scalar projection can be a great aid in understanding how vectors relate in terms of direction and size.
Calculus
Calculus is the mathematical study that focuses on change. Vector calculus extends these principles into multi-dimensional spaces. Understanding how vectors behave in this context is crucial in many fields, such as physics and engineering.
  • Calculus with vectors often involves operations like differentiation and integration in multi-dimensional spaces.
  • This exercise focuses on projections, which can be understood better with a calculus backdrop, especially when dealing with phenomena like field lines or gradients in physical sciences.
In applied contexts, vector calculus helps in representing forces, velocities, or any other vectorial quantities that change over time or space.

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Most popular questions from this chapter

Find the distance between the given parallel planes. \(2 x-3 y+z=4, \quad 4 x-6 y+2 z=3\)

Where does the line through \((1,0,1)\) and \((4,-2,2)\) intersect the plane \(x+y+z=6 ?\)

\(29-36\) Reduce the equation to one of the standard forms, classify the surface, and sketch it. $$4 x^{2}+y^{2}+4 z^{2}-4 y-24 z+36=0$$

If \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) are noncoplanar vectors, let $$\mathbf{k}_{1}=\frac{\mathbf{v}_{2} \times \mathbf{v}_{3}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}$$ $$\mathbf{k}_{2}=\frac{\mathbf{v}_{3} \times \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}$$ $$\mathbf{k}_{3}=\frac{\mathbf{v}_{1} \times \mathbf{v}_{2}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}$$ These vectors occur in the study of crystallography. Vectors of the form \(n_{1} \mathbf{v}_{1}+n_{2} \mathbf{v}_{2}+n_{3} \mathbf{v}_{3},\) where each \(n_{i}\) is an integer, form a lattice for a crystal. Vectors written similarly in terms of \(\mathbf{k}_{1}, \mathbf{k}_{2},\) and \(\mathbf{k}_{3}\) form the reciprocal lattice.) (a) Show that \(\mathbf{k}_{i}\) is perpendicular to \(\mathbf{v},\) if \(i \neq j\) (b) Show that \(\mathbf{k}_{i} \cdot \mathbf{v}_{t}=1\) for \(i=1,2,3 .\) (c) Show that \(\mathbf{k}_{1} \cdot\left(\mathbf{k}_{2} \times \mathbf{k}_{3}\right)=\frac{1}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\)

Find the area of the parallelogram with vertices \(A(-2,1),\) \(B(0,4), C(4,2),\) and \(D(2,-1)\)
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