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Chapter 11: Problem 12

Let \(\mathbf{u}=\langle 1,2\rangle, \mathbf{v}=\langle 4,-2\rangle\), and \(\mathbf{w}=\langle 6,0\rangle .\) Find (a) \(\mathbf{u} \cdot(7 \mathbf{v}+\mathbf{w})\) (b) \(\|(\mathbf{u} \cdot \mathbf{w}) \mathbf{w}\|\) (c) \(\|\mathbf{u}\|(\mathbf{v} \cdot \mathbf{w})\) (d) \((\|\mathbf{u}\| \mathbf{v}) \cdot \mathbf{w}\)

Short Answer

Expert verified
(a) 6 (b) 36 (c) 24\sqrt{5} (d) 24\sqrt{5}

Step by step solution

01

Calculate 7v

First, calculate the vector obtained by multiplying vector \( \mathbf{v} = \langle 4, -2 \rangle \) by 7. Multiply each component of \( \mathbf{v} \) by 7:\[ 7 \mathbf{v} = 7 \times \langle 4, -2 \rangle = \langle 28, -14 \rangle \]
02

Find 7v + w

Next, add the vector \( \mathbf{w} = \langle 6, 0 \rangle \) to the calculated vector \( 7 \mathbf{v} = \langle 28, -14 \rangle \):\[7 \mathbf{v} + \mathbf{w} = \langle 28, -14 \rangle + \langle 6, 0 \rangle = \langle 34, -14 \rangle \]
03

Compute u 路 (7v + w)

Calculate the dot product between \( \mathbf{u} = \langle 1, 2 \rangle \) and \( 7 \mathbf{v} + \mathbf{w} = \langle 34, -14 \rangle \):\[ \mathbf{u} \cdot (7 \mathbf{v} + \mathbf{w}) = 1 \times 34 + 2 \times (-14) = 34 - 28 = 6 \]
04

Calculate u 路 w

Determine the dot product between vectors \( \mathbf{u} = \langle 1, 2 \rangle \) and \( \mathbf{w} = \langle 6, 0 \rangle \):\[ \mathbf{u} \cdot \mathbf{w} = 1 \times 6 + 2 \times 0 = 6 + 0 = 6 \]
05

Find (u 路 w)w

Using the result from \( \mathbf{u} \cdot \mathbf{w} = 6 \), multiply this scalar by the vector \( \mathbf{w} \):\[ (\mathbf{u} \cdot \mathbf{w}) \mathbf{w} = 6 \times \langle 6, 0 \rangle = \langle 36, 0 \rangle \]
06

Compute the magnitude of (u 路 w)w

Find the magnitude of the vector \( \langle 36, 0 \rangle \):\[ \| (\mathbf{u} \cdot \mathbf{w}) \mathbf{w} \| = \sqrt{36^2 + 0^2} = \sqrt{1296} = 36 \]
07

Calculate the magnitude of u

Calculate the magnitude of vector \( \mathbf{u} = \langle 1, 2 \rangle \):\[ \| \mathbf{u} \| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]
08

Compute v 路 w

Calculate the dot product between vectors \( \mathbf{v} = \langle 4, -2 \rangle \) and \( \mathbf{w} = \langle 6, 0 \rangle \):\[ \mathbf{v} \cdot \mathbf{w} = 4 \times 6 + (-2) \times 0 = 24 + 0 = 24 \]
09

Calculate ||u||(v 路 w)

Multiply the magnitude of \( \mathbf{u} \) by the result from the dot product \( \mathbf{v} \cdot \mathbf{w} \):\[ \| \mathbf{u} \| (\mathbf{v} \cdot \mathbf{w}) = \sqrt{5} \times 24 = 24\sqrt{5} \]
10

Compute ||u||v

Multiply the magnitude of \( \mathbf{u} \) \( \sqrt{5} \) with the vector \( \mathbf{v} \):\[ \| \mathbf{u} \| \mathbf{v} = \sqrt{5} \times \langle 4, -2 \rangle = \langle 4\sqrt{5}, -2\sqrt{5} \rangle \]
11

Find (||u||v) 路 w

Calculate the dot product between the vector from \( \| \mathbf{u} \| \mathbf{v} = \langle 4\sqrt{5}, -2\sqrt{5} \rangle \) and \( \mathbf{w} = \langle 6, 0 \rangle \):\[ (\| \mathbf{u} \| \mathbf{v}) \cdot \mathbf{w} = 4\sqrt{5} \times 6 + (-2\sqrt{5}) \times 0 = 24\sqrt{5} \]

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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 vector operation that combines two vectors resulting in a scalar. To calculate it for two vectors \(\mathbf{a}\) and \(\mathbf{b}\), represented as \(\langle a_1, a_2 \rangle\) and \(\langle b_1, b_2 \rangle\) respectively, you multiply the corresponding components and sum the results:
  • Formula: \(\mathbf{a} \cdot \mathbf{b} = a_1 \times b_1 + a_2 \times b_2\)
In our problem, the dot product \(\mathbf{u} \cdot \mathbf{w}\) for vectors \(\mathbf{u}=\langle 1, 2 \rangle\) and \(\mathbf{w}=\langle 6, 0 \rangle\) is calculated as follows:
  • \(\mathbf{u} \cdot \mathbf{w} = 1 \times 6 + 2 \times 0 = 6\)
This operation helps find projections of one vector onto another and plays a crucial role in determining angles between vectors.
Magnitude of a Vector
The magnitude of a vector, also known as the length or norm, is a measure of how long a vector is. Calculating the magnitude involves using the Pythagorean theorem when vectors are represented in a Cartesian plane.
  • Formula: For a vector \(\mathbf{a}=\langle a_1, a_2 \rangle\), the magnitude is \(\|\mathbf{a}\|=\sqrt{a_1^2 + a_2^2}\).
In the exercise, for the vector \(\mathbf{u}=\langle 1, 2 \rangle\), its magnitude is:
  • \(\|\mathbf{u}\| = \sqrt{1^2 + 2^2} = \sqrt{5}\).
A vector's magnitude is always a non-negative number. It is particularly useful in physics and engineering to describe the size or strength of a quantity such as velocity or force.
Scalar Multiplication
Scalar multiplication is a simple yet powerful operation where each component of a vector is multiplied by the same scalar (a real number). This operation scales the vector without changing its direction unless the scalar is negative.
  • Example: If vector \(\mathbf{v} = \langle v_1, v_2 \rangle\) and scalar \(c\), then \(c \mathbf{v} = \langle c \cdot v_1, c \cdot v_2 \rangle\).
In the given problem, multiplying the vector \(\mathbf{v} = \langle 4, -2 \rangle\) by scalar 7 gives:
  • \(7\mathbf{v} = \langle 7 \times 4, 7 \times (-2) \rangle = \langle 28, -14 \rangle\).
Scalar multiplication is essential in vector spaces for scaling vectors, changing their magnitude, and impacting operations like vector addition and dot products.

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Most popular questions from this chapter

Suppose that the temperature \(T\) at a point \((x, y, z)\) on the line \(x=t, y=1+t, z=3-2 t\) is \(T=25 x^{2} y z .\) Use a CAS or a calculating utility with a root-finding capability to approximate the maximum temperature on that portion of the line that extends from the \(x z\) -plane to the \(x y\) -plane.

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