The number of unit vectors perpendicular to the vector `vec(a) = 2 hat(i) + hat(j) + 2 hat(k)` and `vec(b) = hat(j) + hat(k)` is

Options

A. one
B. two
C. three
D. infinite

Answer

To find the number of unit vectors perpendicular to the vectors \(\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Find the Cross Product of \(\vec{a}\) and \(\vec{b}\)

The cross product \(\vec{a} \times \vec{b}\) will give us a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \]
\[ \vec{b} = 0\hat{i} + \hat{j} + \hat{k} \]

Using the determinant method for the cross product:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1\)

2. \(\begin{vmatrix} 2 & 2 \\ 0 & 1 \end{vmatrix} = (2)(1) - (2)(0) = 2 - 0 = 2\)

3. \(\begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} = (2)(1) - (1)(0) = 2 - 0 = 2\)

Putting it all together:

\[ \vec{a} \times \vec{b} = -1\hat{i} - 2\hat{j} + 2\hat{k} = -\hat{i} - 2\hat{j} + 2\hat{k} \]

### Step 2: Find the Magnitude of the Cross Product

Now we find the magnitude of \(\vec{a} \times \vec{b}\):

\[ |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 3: Find the Unit Vectors

The unit vectors perpendicular to both \(\vec{a}\) and \(\vec{b}\) can be found by dividing the cross product by its magnitude:

\[ \text{Unit vector} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{3} \]

This results in:

\[ \text{Unit vector} = -\frac{1}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \]

### Step 4: Consider the Negative Unit Vector

Since the question asks for the number of unit vectors, we also have the negative of the unit vector:

\[ \text{Negative unit vector} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k} \]

### Conclusion

Thus, there are **two unit vectors** that are perpendicular to both \(\vec{a}\) and \(\vec{b}\).

### Final Answer

The number of unit vectors perpendicular to \(\vec{a}\) and \(\vec{b}\) is **2**.

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