Find a unit vector parallel to the xy plane and perpendicular to the vector 4i- 3j +k .

To find a unit vector that is both parallel to the xy-plane and perpendicular to the vector $\mathbf{v}=4\mathbf{i}-3\mathbf{j}+\mathbf{k}$, we need to break down the problem into manageable steps.

Understanding the Concepts

The xy-plane is characterized by vectors that have no component in the z-direction. Therefore, any vector in the xy-plane can be represented as $\mathbf{a}=a\mathbf{i}+b\mathbf{j}$, where $a$ and $b$ are real numbers. To satisfy the condition of being perpendicular to $\mathbf{v}$, we will utilize the dot product, which is defined as follows:

The dot product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is zero if the vectors are perpendicular. Thus, we need:

$\mathbf{a}\cdot \mathbf{v}=0$

Setting Up the Equation

Given $\mathbf{v}=4\mathbf{i}-3\mathbf{j}+\mathbf{k}$, the dot product with $\mathbf{a}=a\mathbf{i}+b\mathbf{j}$ is calculated as follows:

$\mathbf{a}\cdot \mathbf{v}=(a\mathbf{i}+b\mathbf{j})\cdot (4\mathbf{i}-3\mathbf{j}+\mathbf{k})$

This simplifies to:

$4a-3b+0=0$

Finding Relationships Between a and b

From the equation $4a-3b=0$, we can express one variable in terms of the other. Rearranging gives:

$4a=3b$

Thus, we can express $b$ as:

$b=\frac{4}{3}a$

Selecting a Value for a

Now, we can choose a value for $a$ to find a specific vector. Let’s set $a=3$. Plugging this into the equation for $b$ gives:

$b=\frac{4}{3}\cdot 3=4$

So one vector in the xy-plane that is perpendicular to $\mathbf{v}$ is:

$\mathbf{a}=3\mathbf{i}+4\mathbf{j}$

Creating the Unit Vector

To convert $\mathbf{a}$ into a unit vector, we need to find its magnitude:

$||\mathbf{a}||=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

Now, we divide each component of $\mathbf{a}$ by its magnitude:

$\mathbf{u}=\frac{1}{5}(3\mathbf{i}+4\mathbf{j})=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

Final Result

The unit vector that is parallel to the xy-plane and perpendicular to $\mathbf{v}$ is:

$\mathbf{u}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

This vector satisfies both conditions and is a valid unit vector in the xy-plane. You can verify the calculations and the properties of the vectors to deepen your understanding of the concepts involved!