Problem 16 - Essential Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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: \( 2\mathbf{i} + (-4 + 2)\mathbf{j} + (4 - 1)\mathbf{k} = 2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \).

Step 2: 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: \( 4\mathbf{i} + (-8 + 6)\mathbf{j} + (8 - 3)\mathbf{k} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \).

Step 3: 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: \( |\mathbf{a}| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \).

Step 4: 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: \( |\mathbf{a} - \mathbf{b}| = \sqrt{(2)^2 + (-6)^2 + (5)^2} = \sqrt{4 + 36 + 25} = \sqrt{65} \).

Key Concepts

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.