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

Answer

To solve the problem step by step, we will follow the outlined process in the video transcript.

### Step 1: Identify the vectors
We have two vectors:
- \( \mathbf{A} = \hat{i} - \hat{j} + 2\hat{k} \)
- \( \mathbf{B_1} = 2\hat{i} + \hat{j} + 3\hat{k} \)
- \( \mathbf{B_2} = \hat{i} - \hat{j} - 2\hat{k} \)

### Step 2: Find the vector perpendicular to the plane formed by \( \mathbf{B_1} \) and \( \mathbf{B_2} \)
To find a vector perpendicular to the plane formed by \( \mathbf{B_1} \) and \( \mathbf{B_2} \), we can use the cross product:

\[ \mathbf{B} = \mathbf{B_1} \times \mathbf{B_2} \]

Calculating the cross product:

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

Expanding this determinant:

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

Calculating the determinants:

1. For \( \hat{i} \):
\[ 1 \cdot (-2) - 3 \cdot (-1) = -2 + 3 = 1 \]

2. For \( \hat{j} \):
\[ 2 \cdot (-2) - 3 \cdot 1 = -4 - 3 = -7 \quad \text{(note the negative sign)} \]

3. For \( \hat{k} \):
\[ 2 \cdot (-1) - 1 \cdot 1 = -2 - 1 = -3 \]

Putting it all together:

\[ \mathbf{B} = \hat{i} - (-7)\hat{j} - 3\hat{k} = \hat{i} + 7\hat{j} - 3\hat{k} \]

### Step 3: Calculate the projection of \( \mathbf{A} \) on \( \mathbf{B} \)
The formula for the projection of vector \( \mathbf{A} \) onto vector \( \mathbf{B} \) is given by:

\[ \text{Projection of } \mathbf{A} \text{ on } \mathbf{B} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|} \]

Calculating \( \mathbf{A} \cdot \mathbf{B} \):

\[ \mathbf{A} \cdot \mathbf{B} = (1)(1) + (-1)(7) + (2)(-3) = 1 - 7 - 6 = -12 \]

Now, we need the magnitude of \( \mathbf{B} \):

\[ |\mathbf{B}| = \sqrt{1^2 + 7^2 + (-3)^2} = \sqrt{1 + 49 + 9} = \sqrt{59} \]

Thus, the projection of \( \mathbf{A} \) on \( \mathbf{B} \) is:

\[ \text{Projection} = \frac{-12}{\sqrt{59}} \]

### Step 4: Find \( k \) and calculate \( \frac{1}{k^2} \)
Given that the magnitude of the projection is \( k \):

\[ k = \left| \frac{-12}{\sqrt{59}} \right| = \frac{12}{\sqrt{59}} \]

Now, we need to find \( \frac{1}{k^2} \):

\[ k^2 = \left( \frac{12}{\sqrt{59}} \right)^2 = \frac{144}{59} \]

Thus,

\[ \frac{1}{k^2} = \frac{59}{144} \]

### Final Answer
The value of \( \frac{1}{k^2} \) is:

\[ \frac{59}{144} \]