The number of unit vectors perpendicular to the plane of vectors `veca=2hati-6hat j-3hatk` and `vecb=4hati+3hatj-hatk` is/are

Options

A. 1
B. 2
C. 0
D. None of these

Answer

To find the number of unit vectors that are perpendicular to the plane defined by the vectors \(\vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k}\) and \(\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}\), we will follow these steps:

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

The cross product \(\vec{a} \times \vec{b}\) gives us a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\). We can set up the determinant to find the cross product:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{vmatrix} \]

Calculating this determinant, we have:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -6 & -3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} -6 & -3 \\ 3 & -1 \end{vmatrix} = (-6)(-1) - (-3)(3) = 6 + 9 = 15 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} = (2)(-1) - (-3)(4) = -2 + 12 = 10 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} = (2)(3) - (-6)(4) = 6 + 24 = 30 \]

Putting it all together, we have:
\[ \vec{a} \times \vec{b} = 15\hat{i} - 10\hat{j} + 30\hat{k} \]

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

Next, we calculate the magnitude of the cross product vector:
\[ |\vec{a} \times \vec{b}| = \sqrt{15^2 + (-10)^2 + 30^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35 \]

### Step 3: Calculate the Unit Vector

The unit vector \(\hat{n}\) in the direction of \(\vec{a} \times \vec{b}\) is given by:
\[ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \]
Substituting the values:
\[ \hat{n} = \frac{15\hat{i} - 10\hat{j} + 30\hat{k}}{35} = \left(\frac{15}{35}\right)\hat{i} + \left(\frac{-10}{35}\right)\hat{j} + \left(\frac{30}{35}\right)\hat{k} \]
Simplifying this gives:
\[ \hat{n} = \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \]

### Step 4: Determine the Number of Unit Vectors

Since the unit vector can point in both directions (positive and negative), the two unit vectors perpendicular to the plane are:
\[ \hat{n} = \frac{3}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \quad \text{and} \quad -\hat{n} = -\frac{3}{7}\hat{i} + \frac{2}{7}\hat{j} - \frac{6}{7}\hat{k} \]

Thus, the number of unit vectors perpendicular to the plane of vectors \(\vec{a}\) and \(\vec{b}\) is **2**.

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