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Chapter 10: Problem 16

Find \(\mathbf{a}+\mathbf{b}, 2 \mathbf{a}+3 \mathbf{b},|\mathbf{a}|,\) and \(|\mathbf{a}-\mathbf{b}|\) \(\mathbf{a}=2 \mathbf{i}-4 \mathbf{j}+4 \mathbf{k}, \quad \mathbf{b}=2 \mathbf{j}-\mathbf{k}\)

Short Answer

Expert verified
\( \mathbf{a} + \mathbf{b} = 2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}, 2\mathbf{a} + 3\mathbf{b} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}, |\mathbf{a}| = 6, |\mathbf{a} - \mathbf{b}| = \sqrt{65} \).

Step by step solution

01

Calculate \( \mathbf{a} + \mathbf{b} \)

To find \( \mathbf{a} + \mathbf{b} \), we need to add the corresponding components of the vectors. \[ \mathbf{a} = 2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k}, \quad \mathbf{b} = 2 \mathbf{j} - \mathbf{k} \] So, \( \mathbf{a} + \mathbf{b} = (2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}) + (0\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \). Calculating component-wise gives: \( 2\mathbf{i} + (-4 + 2)\mathbf{j} + (4 - 1)\mathbf{k} = 2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \).
02

Calculate \( 2 \mathbf{a} + 3 \mathbf{b} \)

First, find \( 2\mathbf{a} \) by multiplying each component of \( \mathbf{a} = 2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k} \) by 2: \( 2(2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k}) = 4\mathbf{i} - 8\mathbf{j} + 8\mathbf{k} \). Next, find \( 3\mathbf{b} \): \( 3(0\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 0\mathbf{i} + 6\mathbf{j} - 3\mathbf{k} \). Now, add \( 4\mathbf{i} - 8\mathbf{j} + 8\mathbf{k} \) and \( 0\mathbf{i} + 6\mathbf{j} - 3\mathbf{k} \) to get: \( 4\mathbf{i} + (-8 + 6)\mathbf{j} + (8 - 3)\mathbf{k} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \).
03

Calculate \( |\mathbf{a}| \)

The magnitude of \( \mathbf{a} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k} \) is given by \( |\mathbf{a}| = \sqrt{(2)^2 + (-4)^2 + (4)^2} \). Performing the calculations gives: \( |\mathbf{a}| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \).
04

Calculate \( |\mathbf{a} - \mathbf{b}| \)

First, calculate \( \mathbf{a} - \mathbf{b} = (2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k}) - (0 \mathbf{i} + 2 \mathbf{j} - \mathbf{k}) \). Subtracting component-wise results in: \( 2\mathbf{i} + (-4 - 2)\mathbf{j} + (4 + 1)\mathbf{k} = 2\mathbf{i} - 6\mathbf{j} + 5\mathbf{k} \). Now, find the magnitude: \( |\mathbf{a} - \mathbf{b}| = \sqrt{(2)^2 + (-6)^2 + (5)^2} = \sqrt{4 + 36 + 25} = \sqrt{65} \).

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Key Concepts

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

Vector Magnitude
Every vector has a magnitude, which is essentially its length. To determine the magnitude of a vector, we use the formula \[ | extbf{v}| = \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2} \]where \(x_1, x_2,\) and \(x_3\) are the components of the vector.
  • For vector \( \mathbf{a} = 2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k} \), the components are \(2, -4, \) and \(4\).
  • Calculating its magnitude involves squaring each component: \(2^2 + (-4)^2 + 4^2\).
  • This results in \( 4 + 16 + 16 = 36\).
  • The square root of \(36\) is \(6\), so \(|\mathbf{a}| = 6\).
Knowing the magnitude is crucial because it allows us to understand the size of the vector in the space it resides.
It's similar to finding the length of a line segment in geometry.
Vector Subtraction
Vector subtraction is the process of finding the vector difference between two vectors. This is achieved by subtracting each component of one vector from the corresponding component of another. The operation is represented as:\[ \mathbf{a} - \mathbf{b} = (a_1 - b_1) \, \mathbf{i} + (a_2 - b_2) \, \mathbf{j} + (a_3 - b_3) \, \mathbf{k} \]
  • Given vectors \(\mathbf{a} = 2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k} \) and \(\mathbf{b} = 0 \mathbf{i} + 2 \mathbf{j} - \mathbf{k} \), apply the formula.
  • Subtract component-wise: \(2 - 0, \; -4 - 2, \; 4 - (-1)\).
  • This results in the vector \(2 \mathbf{i} - 6 \mathbf{j} + 5 \mathbf{k} \).
Calculating vector subtraction gives us a new vector that shows both the direction and distance of the difference between \( \mathbf{a} \) and \( \mathbf{b} \). This can be useful in physics for determining change or displacement.
Scalar Multiplication of Vectors
Scalar multiplication involves multiplying a vector by a scalar (a real number). The process scales the vector, changing its magnitude but not its direction.
  • For vector \( \mathbf{a} = 2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k} \), multiplying by scalar \(2\) yields \( 2 \times (2 \mathbf{i} - 4 \mathbf{j} + 4 \mathbf{k}) = 4 \mathbf{i} - 8 \mathbf{j} + 8 \mathbf{k} \).
  • The vector \(\mathbf{b} = 0\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) multiplied by scalar \(3\) becomes \(3 \times (2 \mathbf{j} - \mathbf{k}) = 6 \mathbf{j} - 3 \mathbf{k} \).
  • Scalar multiplication affects only the size of the vector, not its direction, unless multiplied by a negative, which reverses its direction.
The result in any scalar multiplication is another vector, making scalar multiplication a straightforward way to alter a vector's scale.
This concept is notably important in physics, where vectors often need to be scaled in calculations.

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Most popular questions from this chapter

What force is required so that a particle of mass \(m\) has the position function \(\mathbf{r}(t)=t^{3} \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} ?\)

\(45-46\) . Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t\) . $$\mathbf{r}(t)=\left\langle t^{3}+3 t, t^{2}+1,3 t+4\right\rangle, \quad t=1$$

Let \(L_{1}\) be the line through the points \((1,2,6)\) and \((2,4,8) .\) Let \(L_{2}\) be the line of intersection of the planes \(\pi_{1}\) and \(\pi_{2}\) where \(\pi_{1}\) is the plane \(x-y+2 z+1=0\) and \(\pi_{2}\) is the plane through the points \((3,2,-1),(0,0,1),\) and \((1,2,1)\) . Calculate the distance between \(L_{1}\) and \(L_{2}\) .

\(49-52=\) Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$x=e^{-t} \cos t, \quad y=e^{-t} \sin t, \quad z=e^{-t_{ ;}},(1,0,1)$$

(a) Find the point of intersection of the tangent lines to the curve \(\mathbf{r}(t)=\langle\sin \pi t, 2 \sin \pi t, \cos \pi t\rangle\) at the points where \( t=0\) and \(t=0.5\) . (b) Illustrate by graphing the curve and both tangent lines.
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