Question

Let \(\mathbf{A}=2 \mathbf{i}-\mathbf{j}+2 \mathbf{k} .\) (a) Find a unit vector in the same direction as \(\mathbf{A}\). Hint: Divide A by | Al. (b) Find a vector in the same direction as A but of magnitude 12 . (c) Find a vector perpendicular to A. Hint: There are many such vectors; you are to find one of them. (d) Find a unit vector perpendicular to A. See hint in (a).

Short answer

The unit vector in the same direction as \( \mathbf{A} \) is \( \frac{2}{3} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{2}{3} \mathbf{k} \). The vector with magnitude 12 in the same direction is \( 8 \mathbf{i} - 4 \mathbf{j} + 8 \mathbf{k} \). One vector perpendicular to \( \mathbf{A} \) is \( \mathbf{j} + \mathbf{k} \). The unit vector perpendicular to \( \mathbf{A} \) is \( \frac{1}{\sqrt{2}} \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k} \).

Step-by-step solution

Find the Magnitude of \mathbf{A}

First, calculate the magnitude of \( \mathbf{A} \). The vector \( \mathbf{A} \) is given by \( \mathbf{A} = 2 \mathbf{i} - \mathbf{j} + 2 \mathbf{k} \). The magnitude \( |\mathbf{A}| \) is found using the formula \[ |\mathbf{A}| = \sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Find the Unit Vector in the Direction of \mathbf{A}

To find the unit vector in the same direction as \( \mathbf{A} \), divide \( \mathbf{A} \) by its magnitude \( | \mathbf{A} | \): \[ \hat{A} = \frac{\mathbf{A}}{|\mathbf{A}|} = \frac{2 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}}{3} = \frac{2}{3} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{2}{3} \mathbf{k} \]

Find a Vector in the Same Direction as \mathbf{A} with Magnitude 12

To find a vector in the same direction as \( \mathbf{A} \) but with a different magnitude, multiply the unit vector \( \hat{A} \) by the desired magnitude: \[ \mathbf{B} = 12 \hat{A} = 12 \left( \frac{2}{3} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{2}{3} \mathbf{k} \right) = 8 \mathbf{i} - 4 \mathbf{j} + 8 \mathbf{k} \]

Find a Vector Perpendicular to \mathbf{A}

A simple approach to find a vector perpendicular to \( \mathbf{A} \) is to choose any vector that satisfies the dot product condition \( \mathbf{A} \cdot \mathbf{C} = 0 \). Let \( \mathbf{C} = \mathbf{i} + \mathbf{j} \). Now, check the dot product: \[ \mathbf{A} \cdot \mathbf{C} = (2 \mathbf{i} - \mathbf{j} + 2 \mathbf{k}) \cdot (\mathbf{i} + \mathbf{j}) = 2 \cdot 1 + (-1) \cdot 1 + 2 \cdot 0 = 2 - 1 = 1 \] So choose another vector, let \( \mathbf{C} = \mathbf{j} + \mathbf{k} \): \[ \mathbf{A} \cdot \mathbf{C} = 2 \cdot 0 + (-1) \cdot 1 + 2 \cdot 1 = 0 \] Thus, \( \mathbf{C} = \mathbf{j} + \mathbf{k} \) is perpendicular to \( \mathbf{A} \).

Find a Unit Vector Perpendicular to \mathbf{A}

To find a unit vector perpendicular to \( \mathbf{A} \), normalize the perpendicular vector \( \mathbf{C} \) found in the previous step. The magnitude \( |\mathbf{C}| \) of \( \mathbf{C} = \mathbf{j} + \mathbf{k} \) is \[ |\mathbf{C}| = \sqrt{(0)^2 + (1)^2 + (1)^2} = \sqrt{2} \] Therefore, the unit vector is \[ \hat{C} = \frac{\mathbf{C}}{|\mathbf{C}} = \frac{\mathbf{j} + \mathbf{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \mathbf{j} + \frac{1}{\sqrt{2}} \mathbf{k}\]

Key concepts

Unit Vector

To understand what a unit vector is, let's break it down simply. A unit vector is a vector with a magnitude of 1. Unit vectors are useful because they express direction without regard for magnitude.
To find a unit vector in the same direction as a given vector \(\mathbf{A}\), you need to divide each component of \(\mathbf{A}\) by its magnitude.
In the problem, we have the vector \(\mathbf{A} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}.\) Its magnitude is calculated as:
\[|\mathbf{A}| = \sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\]
Now, to get the unit vector \(\hat{A}\), we divide each component of \(\mathbf{A}\) by its magnitude:
\[\hat{A} = \frac{2\mathbf{i} - \mathbf{j} + 2\mathbf{k}}{3} = \frac{2}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}\]
So, \(\hat{A}\) is the unit vector in the direction of \(\mathbf{A}\).

Vector Magnitude

The magnitude of a vector tells us how long the vector is. It’s also called the vector length or norm. The magnitude of a vector \(\mathbf{A} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) is calculated using the formula:
\[|\mathbf{A}| = \sqrt{a^2 + b^2 + c^2}\]
This formula comes from the Pythagorean theorem.
In our example, the vector \(\mathbf{A}\) is \(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\). The magnitude is:
\[|\mathbf{A}| = \sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\]
Understanding the magnitude makes it easier to scale vectors and also helps in understanding unit vector calculations.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors, producing a scalar (a single number).
For vectors \(\mathbf{A}\) and \(\mathbf{B}\): \[\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3\] where \(\mathbf{A} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{B} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\).
The dot product is useful in determining the angle between two vectors or if they are perpendicular.
For vectors to be perpendicular, their dot product must be zero. In our example, we checked the vector \(\mathbf{j} + \mathbf{k}\) with \(\mathbf{A}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}\) and found:
\[\mathbf{A} \cdot (\mathbf{j} + \mathbf{k}) = 2 \cdot 0 + (-1) \cdot 1 + 2 \cdot 1 = 0 \]
This confirmed that \(\mathbf{j} + \mathbf{k}\) is indeed perpendicular to \(\mathbf{A}\).

Perpendicular Vectors

Vectors are considered perpendicular if the angle between them is 90 degrees. Mathematically, this means their dot product is zero.
Given a vector \(\mathbf{A}\), finding a perpendicular vector \(\mathbf{C}\) involves choosing any vector that satisfies \(\mathbf{A} \cdot \mathbf{C} = 0\).
In the problem, we initially tried \(\mathbf{i} + \mathbf{j}\), but it didn't work. We then tried \(\mathbf{j} + \mathbf{k}\):
\