The magnitude of the projection of the vector `2hati+3hatj+hatk` on the vector perpendicular to the plane containing the vectors `hati+hatj+hatk" and "hati+2hatj+3hatk`, is

Options

A. `sqrt((3)/(2))`
B. `3sqrt(6)`
C. `(sqrt(3))/(2)`
D. `sqrt(6)`

Explanation

To solve the problem, we need to find the magnitude of the projection of the vector \( \mathbf{A} = 2\hat{i} + 3\hat{j} + \hat{k} \) onto the normal vector of the plane formed by the vectors \( \mathbf{B} = \hat{i} + \hat{j} + \hat{k} \) and \( \mathbf{C} = \hat{i} + 2\hat{j} + 3\hat{k} \).

### Step 1: Find the normal vector to the plane
The normal vector \( \mathbf{N} \) to the plane can be found using the cross product of vectors \( \mathbf{B} \) and \( \mathbf{C} \).

\[ \mathbf{N} = \mathbf{B} \times \mathbf{C} \]

Calculating the cross product:

\[ \mathbf{B} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{C} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \]

Using the determinant method for the cross product:

\[ \mathbf{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{N} = \hat{i} \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:

\[ \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 1 \]
\[ \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = (1)(3) - (1)(1) = 2 \]
\[ \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 1 \]

Thus, we have:

\[ \mathbf{N} = \hat{i}(1) - \hat{j}(2) + \hat{k}(1) = \hat{i} - 2\hat{j} + \hat{k} \]

### Step 2: Find the magnitude of the normal vector
The magnitude of the normal vector \( \mathbf{N} \) is given by:

\[ |\mathbf{N}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Step 3: Find the projection of \( \mathbf{A} \) onto \( \mathbf{N} \)
The projection of vector \( \mathbf{A} \) onto vector \( \mathbf{N} \) is given by the formula:

\[ \text{proj}_{\mathbf{N}} \mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{N}}{|\mathbf{N}|^2} \mathbf{N} \]

Calculating the dot product \( \mathbf{A} \cdot \mathbf{N} \):

\[ \mathbf{A} \cdot \mathbf{N} = (2\hat{i} + 3\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 2(1) + 3(-2) + 1(1) = 2 - 6 + 1 = -3 \]

Now, substituting into the projection formula:

\[ \text{proj}_{\mathbf{N}} \mathbf{A} = \frac{-3}{|\mathbf{N}|^2} \mathbf{N} = \frac{-3}{6} \mathbf{N} = -\frac{1}{2} \mathbf{N} \]

### Step 4: Find the magnitude of the projection
The magnitude of the projection is:

\[ \left| \text{proj}_{\mathbf{N}} \mathbf{A} \right| = \left| -\frac{1}{2} \mathbf{N} \right| = \frac{1}{2} |\mathbf{N}| = \frac{1}{2} \sqrt{6} \]

### Final Answer
Thus, the magnitude of the projection of the vector \( 2\hat{i} + 3\hat{j} + \hat{k} \) on the vector perpendicular to the plane is:

\[ \frac{\sqrt{6}}{2} \]