A unit vector perpendicular to `3i+4j" and "i-j+k` is

Options

A. `1/sqrt(3)(i+j+k)`
B. `1/sqrt(14)(i-2k+3k)`
C. `1/sqrt(74)(4i+3j-7k)`
D. `1/sqrt(74)(4i-3j-7k)`

Answer

To find a unit vector that is perpendicular to the vectors \( \mathbf{A} = 3\mathbf{i} + 4\mathbf{j} \) and \( \mathbf{B} = \mathbf{i} - \mathbf{j} + \mathbf{k} \), we can follow these steps:

### Step 1: Find the Cross Product of the Two Vectors
The cross product \( \mathbf{A} \times \mathbf{B} \) will give us a vector that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \).

Given:
\[ \mathbf{A} = 3\mathbf{i} + 4\mathbf{j} + 0\mathbf{k} \]
\[ \mathbf{B} = 1\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} \]

We can set up the determinant to find the cross product:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 0 \\ 1 & -1 & 1 \end{vmatrix} \]

### Step 2: Calculate the Determinant
Calculating the determinant, we have:
\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} 4 & 0 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 0 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 4 \\ 1 & -1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 4 & 0 \\ -1 & 1 \end{vmatrix} = (4)(1) - (0)(-1) = 4 \)
2. \( \begin{vmatrix} 3 & 0 \\ 1 & 1 \end{vmatrix} = (3)(1) - (0)(1) = 3 \)
3. \( \begin{vmatrix} 3 & 4 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (4)(1) = -3 - 4 = -7 \)

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 4\mathbf{i} - 3\mathbf{j} - 7\mathbf{k} \]

### Step 3: Find the Magnitude of the Cross Product
Next, we find the magnitude of the vector \( \mathbf{P} = 4\mathbf{i} - 3\mathbf{j} - 7\mathbf{k} \):
\[ |\mathbf{P}| = \sqrt{(4)^2 + (-3)^2 + (-7)^2} = \sqrt{16 + 9 + 49} = \sqrt{74} \]

### Step 4: Find the Unit Vector
To find the unit vector \( \mathbf{p} \) in the direction of \( \mathbf{P} \), we divide \( \mathbf{P} \) by its magnitude:
\[ \mathbf{p} = \frac{\mathbf{P}}{|\mathbf{P}|} = \frac{4\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}}{\sqrt{74}} \]

### Final Answer
Thus, the unit vector perpendicular to \( 3\mathbf{i} + 4\mathbf{j} \) and \( \mathbf{i} - \mathbf{j} + \mathbf{k} \) is:
\[ \mathbf{p} = \frac{4}{\sqrt{74}}\mathbf{i} - \frac{3}{\sqrt{74}}\mathbf{j} - \frac{7}{\sqrt{74}}\mathbf{k} \]

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