The number of vectors of unit length perpendicular to vector `veca-=(5, 6, 0) and vecb-=(6, 5, 0)` is

Options

A. 1
B. 4
C. 3
D. 2

Answer

To solve the problem of finding the number of unit vectors that are perpendicular to the vectors \(\vec{a} = (5, 6, 0)\) and \(\vec{b} = (6, 5, 0)\), we can follow these steps:

### Step 1: Calculate the Cross Product
The first step is to calculate the cross product of the two vectors \(\vec{a}\) and \(\vec{b}\). The cross product \(\vec{a} \times \vec{b}\) will give us a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 6 & 0 \\ 6 & 5 & 0 \end{vmatrix} \]

### Step 2: Expand the Determinant
Now, we will expand the determinant:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 6 & 0 \\ 5 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 0 \\ 6 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & 6 \\ 6 & 5 \end{vmatrix} \]

Calculating these determinants:

- For \(\hat{i}\): \(6 \cdot 0 - 5 \cdot 0 = 0\)
- For \(\hat{j}\): \(5 \cdot 0 - 6 \cdot 0 = 0\)
- For \(\hat{k}\): \(5 \cdot 5 - 6 \cdot 6 = 25 - 36 = -11\)

Thus, we have:

\[ \vec{a} \times \vec{b} = 0\hat{i} - 0\hat{j} - 11\hat{k} = -11\hat{k} \]

### Step 3: Find the Magnitude of the Cross Product
Next, we find the magnitude of the resulting vector \(\vec{n} = -11\hat{k}\):

\[ |\vec{n}| = \sqrt{(-11)^2} = 11 \]

### Step 4: Calculate the Unit Vector
To find the unit vector in the direction of \(\vec{n}\), we divide \(\vec{n}\) by its magnitude:

\[ \text{Unit Vector} = \frac{\vec{n}}{|\vec{n}|} = \frac{-11\hat{k}}{11} = -\hat{k} \]

### Step 5: Identify the Two Unit Vectors
Since a unit vector can point in two opposite directions, we have:

1. \(-\hat{k}\)
2. \(+\hat{k}\)

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

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

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