Consider two vectors  a=−3i−j
  and  b=3i+j+k
 . If  c=xi+yj+kz,x>0
  is the unit vector that is perpendicular to both  a
  and  b
 , find  x,y
  and  z
 . (Write in exact form

To find the unit vector **c** that is perpendicular to both vectors **a** and **b**, we first need to find the cross product of **a** and **b**. The cross product will give us a vector that is perpendicular to both **a** and **b**.

Given:

\[
\mathbf{a} = -3\mathbf{i} - \mathbf{j}
\]
\[
\mathbf{b} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}
\]

### Step 1: Compute the Cross Product \(\mathbf{a} \times \mathbf{b}\)

The cross product of two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is given by the determinant:

\[
\mathbf{a} \times \mathbf{b} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-3 & -1 & 0 \\
3 & 1 & 1
\end{vmatrix}
\]

We can calculate this determinant:

\[
\mathbf{a} \times \mathbf{b} = \mathbf{i} \left( (-1)(1) - (0)(1) \right) - \mathbf{j} \left( (-3)(1) - (0)(3) \right) + \mathbf{k} \left( (-3)(1) - (3)(-1) \right)
\]

Simplifying each component:

\[
\mathbf{a} \times \mathbf{b} = \mathbf{i}(-1) - \mathbf{j}(-3) + \mathbf{k}(-3 + 3)
\]

\[
\mathbf{a} \times \mathbf{b} = -\mathbf{i} + 3\mathbf{j} + 0\mathbf{k}
\]

Thus, the cross product is:

\[
\mathbf{a} \times \mathbf{b} = -\mathbf{i} + 3\mathbf{j}
\]

### Step 2: Convert the Cross Product to a Unit Vector

Now, to find the unit vector \(\mathbf{c}\) in the same direction as \(\mathbf{a} \times \mathbf{b}\), we first compute the magnitude of \(\mathbf{a} \times \mathbf{b}\):

\[
\|\mathbf{a} \times \mathbf{b}\| = \sqrt{(-1)^2 + (3)^2 + 0^2} = \sqrt{1 + 9} = \sqrt{10}
\]

The unit vector \(\mathbf{c}\) in the direction of \(\mathbf{a} \times \mathbf{b}\) is:

\[
\mathbf{c} = \frac{\mathbf{a} \times \mathbf{b}}{\|\mathbf{a} \times \mathbf{b}\|}
\]

Substitute the values:

\[
\mathbf{c} = \frac{-\mathbf{i} + 3\mathbf{j}}{\sqrt{10}}
\]

### Step 3: Identify \(x, y,\) and \(z\)

Since \(\mathbf{c} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\), we have:

\[
x = -\frac{1}{\sqrt{10}}, \quad y = \frac{3}{\sqrt{10}}, \quad z = 0
\]

Thus, the exact values of \(x, y,\) and \(z\) are:

\[
x = -\frac{1}{\sqrt{10}}, \quad y = \frac{3}{\sqrt{10}}, \quad z = 0
\]

Since \(x > 0\), the correct unit vector is the negative of this vector:

\[
x = \frac{1}{\sqrt{10}}, \quad y = -\frac{3}{\sqrt{10}}, \quad z = 0
\]

### Final Answer:

\[
x = \frac{1}{\sqrt{10}}, \quad y = -\frac{3}{\sqrt{10}}, \quad z = 0
\]

Would you like more details, or do you have any questions? 

### Related Questions:

1. What is a cross product and how is it used in vector analysis?
2. How do you find the magnitude of a vector?
3. What are unit vectors, and why are they important?
4. How can you determine if two vectors are perpendicular?
5. How does the cross product relate to the area of a parallelogram formed by two vectors?

### Tip:

When working with vectors, visualizing them in a 3D coordinate system can help you better understand their directions and relationships.

Learn how to find the unit vector perpendicular to vectors a = -3i - j and b = 3i + j + k with a detailed step-by-step solution. This problem involves calculating the cross product of two vectors and determining the unit vector in exact form. Suitable for advanced high school students.

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