If the magnitude of the projection of the vector `hati-hatj+2hatk` on the vector perpendicular to the plane containing the vectors `2hati0hatj+3hatk and hati-hatj-2hatk` is k, then the value of `(1)/(k^(2))` is equal to

Answer

To solve the problem, we need to find the value of \( \frac{1}{k^2} \), where \( k \) is the magnitude of the projection of the vector \( \mathbf{A} = \hat{i} - \hat{j} + 2\hat{k} \) on the vector \( \mathbf{B} \), which is perpendicular to the plane formed by the vectors \( \mathbf{C} = 2\hat{i} + 0\hat{j} + 3\hat{k} \) and \( \mathbf{D} = \hat{i} - \hat{j} - 2\hat{k} \).

### Step 1: Find the normal vector to the plane
To find the vector \( \mathbf{B} \) that is perpendicular to the plane formed by \( \mathbf{C} \) and \( \mathbf{D} \), we can use the cross product:

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

Calculating the cross product:
\[ \mathbf{C} = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \]

Using the determinant to find the cross product:
\[ \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 1 & -1 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{B} = \hat{i} \begin{vmatrix} 0 & 3 \\ -1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 0 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the minors:
\[ \begin{vmatrix} 0 & 3 \\ -1 & -2 \end{vmatrix} = (0)(-2) - (3)(-1) = 3 \]
\[ \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (3)(1) = -4 - 3 = -7 \]
\[ \begin{vmatrix} 2 & 0 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (0)(1) = -2 \]

Thus,
\[ \mathbf{B} = 3\hat{i} + 7\hat{j} - 2\hat{k} \]

### Step 2: Calculate the projection of \( \mathbf{A} \) onto \( \mathbf{B} \)
The formula for the projection of vector \( \mathbf{A} \) onto vector \( \mathbf{B} \) is given by:

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

First, we need to calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \):
\[ \mathbf{A} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 3 \\ 7 \\ -2 \end{pmatrix} \]
\[ \mathbf{A} \cdot \mathbf{B} = (1)(3) + (-1)(7) + (2)(-2) = 3 - 7 - 4 = -8 \]

Next, we find the magnitude squared of \( \mathbf{B} \):
\[ |\mathbf{B}|^2 = 3^2 + 7^2 + (-2)^2 = 9 + 49 + 4 = 62 \]

Thus, the projection is:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{-8}{62} \mathbf{B} = \frac{-4}{31} \mathbf{B} \]

### Step 3: Find the magnitude of the projection
The magnitude of the projection is:
\[ k = \left| \text{proj}_{\mathbf{B}} \mathbf{A} \right| = \left| \frac{-4}{31} \right| |\mathbf{B}| \]

Calculating \( |\mathbf{B}| \):
\[ |\mathbf{B}| = \sqrt{62} \]

Thus,
\[ k = \frac{4}{31} \sqrt{62} \]

### Step 4: Calculate \( \frac{1}{k^2} \)
Now we compute \( k^2 \):
\[ k^2 = \left( \frac{4}{31} \sqrt{62} \right)^2 = \frac{16 \cdot 62}{31^2} = \frac{992}{961} \]

Finally, we find \( \frac{1}{k^2} \):
\[ \frac{1}{k^2} = \frac{961}{992} \]

### Final Answer
Thus, the value of \( \frac{1}{k^2} \) is \( \frac{961}{992} \).