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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 \(\begin{array}{ll}{\text { (a) } \mathbf{u} \cdot(7 \mathbf{v}+\mathbf{w})} & {\text { (b) }\|(\mathbf{u} \cdot \mathbf{w}) \mathbf{w}\|} \\ {\text { (c) }\|\mathbf{u}\|(\mathbf{v} \cdot \mathbf{w})} & {\text { (d) }(\|\mathbf{u}\| \mathbf{v}) \cdot \mathbf{w}}\end{array}\)

Short Answer

Expert verified
(a) 6, (b) 36, (c) 24鈭�5, (d) 24鈭�5.

Step by step solution

01

Calculate 7v + w

First, compute the vector expression \( 7\mathbf{v} + \mathbf{w} \). Multiply the vector \( \mathbf{v}=\langle 4,-2 \rangle \) by 7, which gives \( 7\mathbf{v} = \langle 28, -14 \rangle \). Then, add \( \mathbf{w} = \langle 6, 0 \rangle \) to get \( 7\mathbf{v} + \mathbf{w} = \langle 28+6, -14+0 \rangle = \langle 34, -14 \rangle \).
02

Compute u 路 (7v + w)

Calculate the dot product \( \mathbf{u} \cdot (7\mathbf{v} + \mathbf{w}) \) where \( \mathbf{u} = \langle 1, 2 \rangle \) and \( 7\mathbf{v} + \mathbf{w} = \langle 34, -14 \rangle \). The dot product is given by \( 1 \times 34 + 2 \times (-14) = 34 - 28 = 6 \).
03

Calculate u 路 w

Find the dot product \( \mathbf{u} \cdot \mathbf{w} \), where \( \mathbf{w} = \langle 6, 0 \rangle \). This is \( 1 \times 6 + 2 \times 0 = 6 \).
04

Calculate (u 路 w) w

Use the result from Step 3 to find \((\mathbf{u} \cdot \mathbf{w}) \mathbf{w}\). Multiply the scalar \( 6 \) by vector \( \mathbf{w} = \langle 6, 0 \rangle \), giving \( 6 \cdot \langle 6, 0 \rangle = \langle 36, 0 \rangle \).
05

Calculate \|(u 路 w) w\|

The magnitude of \( \langle 36, 0 \rangle \) is found using \( \sqrt{36^2 + 0^2} = 36 \).
06

Calculate v 路 w

Now, compute \( \mathbf{v} \cdot \mathbf{w} \) where \( \mathbf{v} = \langle 4, -2 \rangle \) and \( \mathbf{w} = \langle 6, 0 \rangle \). The dot product is \( 4 \times 6 + (-2) \times 0 = 24 \).
07

Calculate \|u\|

The magnitude of \( \mathbf{u} = \langle 1, 2 \rangle \) is given by \( \sqrt{1^2 + 2^2} = \sqrt{5} \).
08

Calculate \|u\|(v 路 w)

Combine results from Steps 6 and 7 to compute \( \|\mathbf{u}\|(\mathbf{v} \cdot \mathbf{w}) \), which is \( \sqrt{5} \times 24 = 24\sqrt{5} \).
09

Multiply \|u\| and v

Compute \( \|\mathbf{u}\| \mathbf{v} \) using the magnitude from Step 7. Multiply \( \sqrt{5} \) by vector \( \mathbf{v} = \langle 4, -2 \rangle \), resulting in \( \langle 4\sqrt{5}, -2\sqrt{5} \rangle \).
10

Calculate (\|u\| v) 路 w

Finally, find the dot product \( (\|\mathbf{u}\| \mathbf{v}) \cdot \mathbf{w} \) where \( \mathbf{w} = \langle 6, 0 \rangle \). This gives \( 4\sqrt{5} \times 6 + (-2\sqrt{5}) \times 0 = 24\sqrt{5} \).

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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, also known as the scalar product, is a fundamental operation in vector calculus. It takes two vectors and returns a scalar value. Imagine you have two vectors, \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \). The dot product is calculated as:
  • Multiply the corresponding components of the vectors: \( a_1 \times b_1 \) and \( a_2 \times b_2 \).
  • Add these products: \( a_1 \times b_1 + a_2 \times b_2 \).
For example, for vectors \( \mathbf{u} = \langle 1, 2 \rangle \) and \( \mathbf{w} = \langle 6, 0 \rangle \), the dot product would be \( 1 \times 6 + 2 \times 0 = 6 \). This operation shows the degree of alignment between the vectors and is often used in physics and engineering. If two vectors are perpendicular, their dot product is zero.
Magnitude of a Vector
The magnitude of a vector refers to its length or size. It is a measure of how far the vector extends in space. For a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), the magnitude, denoted as \( \| \mathbf{v} \| \), is calculated using the Pythagorean theorem:
  • Square each component: \( v_1^2 \) and \( v_2^2 \).
  • Add these squares together: \( v_1^2 + v_2^2 \).
  • Take the square root of the sum: \( \sqrt{v_1^2 + v_2^2} \).
For example, if \( \mathbf{u} = \langle 1, 2 \rangle \), its magnitude is \( \sqrt{1^2 + 2^2} = \sqrt{5} \). Understanding the magnitude helps in identifying how strong or intense a vector is, which is crucial in fields like physics where vectors represent forces.
Scalar Multiplication
Scalar multiplication involves taking a vector and multiplying it by a scalar (a real number). This operation scales the vector, either enlarging or shrinking it without changing its direction.
  • Multiply each component of the vector by the scalar.
For instance, if you have a vector \( \mathbf{v} = \langle 4, -2 \rangle \) and you multiply it by 7, you perform:
  • \( 7 \times 4 = 28 \)
  • \( 7 \times -2 = -14 \)
Thus, \( 7 \mathbf{v} = \langle 28, -14 \rangle \). Scalar multiplication is useful for various applications, especially in stretching vectors to different magnitudes in mechanics and physics.

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Most popular questions from this chapter

Show that the lines \(L_{1}\) and \(L_{2}\) are parallel, and find the distance between them. $$ \begin{array}{l}{L_{1}: x=2-t, \quad y=2 t, \quad z=1+t} \\ {L_{2}: x=1+2 t, \quad y=3-4 t, \quad z=5-2 t}\end{array} $$

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