Finding the Unit Vector Perpendicular to Two Vectors

To find a unit vector that is perpendicular to two given vectors, we first need to find a vector that is perpendicular to both. A common method to find such a vector is by taking the cross product of the two vectors.

Let the two given vectors be $\vec{A} = \vec{a} + \vec{b}$ and $\vec{B} = \vec{a} - \vec{b}$.

We are given the vectors $\vec{a}$ and $\vec{b}$ as:

• $\vec{a} = \hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$

Step 1: Calculate the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$

First, let's find the sum of the vectors $\vec{a}$ and $\vec{b}$:

$\vec{A} = \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k})$

To add vectors, we add their corresponding components:

$\vec{A} = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

Next, let's find the difference between the vectors $\vec{a}$ and $\vec{b}$:

$\vec{B} = \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

To subtract vectors, we subtract their corresponding components:

$\vec{B} = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} - 1\hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k}$

So, the two vectors we need to find a perpendicular unit vector for are $\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{B} = -\hat{j} - 2\hat{k}$.

Step 2: Calculate the Cross Product of the Two Vectors

The cross product $\vec{A} \times \vec{B}$ gives a vector that is perpendicular to both $\vec{A}$ and $\vec{B}$.

The cross product can be calculated using a determinant:

$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}$

Expanding the determinant:

$ = \hat{i}((3)(-2) - (4)(-1)) - \hat{j}((2)(-2) - (4)(0)) + \hat{k}((2)(-1) - (3)(0))$

$ = \hat{i}(-6 - (-4)) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0)$

$ = \hat{i}(-6 + 4) - \hat{j}(-4) + \hat{k}(-2)$

$ = -2\hat{i} + 4\hat{j} - 2\hat{k}$

Let $\vec{N} = -2\hat{i} + 4\hat{j} - 2\hat{k}$. This vector $\vec{N}$ is perpendicular to both $\vec{A}$ and $\vec{B}$.

Step 3: Find the Unit Vector

A unit vector has a magnitude of 1. To find the unit vector in the direction of $\vec{N}$, we divide $\vec{N}$ by its magnitude.

The magnitude of $\vec{N}$ is given by $|\vec{N}| = \sqrt{(-2)^2 + (4)^2 + (-2)^2}$.

$|\vec{N}| = \sqrt{4 + 16 + 4} = \sqrt{24}$

We can simplify $\sqrt{24}$ as $\sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}$.

So, the magnitude is $|\vec{N}| = 2\sqrt{6}$.

The unit vector $\hat{n}$ in the direction of $\vec{N}$ is $\frac{\vec{N}}{|\vec{N}|}$:

$\hat{n} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}}$

Divide each component by the magnitude:

$\hat{n} = \frac{-2}{2\sqrt{6}}\hat{i} + \frac{4}{2\sqrt{6}}\hat{j} + \frac{-2}{2\sqrt{6}}\hat{k}$

Simplify the fractions:

$\hat{n} = \frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} + \frac{-1}{\sqrt{6}}\hat{k}$

This is the unit vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$.

Comparing with Options

Let's compare our calculated unit vector with the given options:

• Option 1: $(2/\sqrt{6}) \hat{i} + (1/\sqrt{6}) \hat{j} + (2/\sqrt{6}) \hat{k}$
• Option 2: $(-1/\sqrt{6}) \hat{i} + (1/\sqrt{6}) \hat{j} - (1/\sqrt{6}) \hat{k}$
• Option 3: $(-1/\sqrt{6}) \hat{i} + (2/\sqrt{6}) \hat{j} - (2/\sqrt{6}) \hat{k}$
• Option 4: $(-1/\sqrt{6}) \hat{i} + (2/\sqrt{6}) \hat{j} - (1/\sqrt{6}) \hat{k}$

Our calculated unit vector is $\frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} + \frac{-1}{\sqrt{6}}\hat{k}$, which matches Option 4.

Revision Table: Key Vector Concepts

Additional Information on Perpendicular Vectors and Cross Product

The cross product is a fundamental operation in vector algebra, particularly useful in physics and engineering to find vectors that are orthogonal (perpendicular) to a plane or a direction determined by two other vectors. The magnitude of the cross product $|\vec{A} \times \vec{B}|$ is equal to the area of the parallelogram formed by vectors $\vec{A}$ and $\vec{B}$. The direction of $\vec{A} \times \vec{B}$ is given by the right-hand rule.

When a question asks for "a unit vector perpendicular to" two vectors, there are actually two such unit vectors, pointing in opposite directions. If $\hat{n}$ is a unit vector perpendicular to two vectors, then $-\hat{n}$ is also a unit vector perpendicular to them. The cross product $\vec{A} \times \vec{B}$ gives one such vector, and $\vec{B} \times \vec{A}$ gives the other, which is the negative of the first. The options provided in a multiple-choice question will typically include only one of these possibilities.

Calculating the cross product involves a determinant calculation, which can sometimes be prone to sign errors. It's important to carefully follow the expansion of the determinant, paying attention to the signs for the $\hat{j}$ term.

The dot product is another important operation. While the cross product gives a perpendicular vector, the dot product tells us about the projection of one vector onto another and can be used to check if two vectors are perpendicular (their dot product is zero).