Let `vec(a) = hat(i) + hat(j) + hat(k), vec(b) and vec(c )= hat(j)-hat(k)` be three vectors such that `vec(a) xx vec(b)= vec(c ) and vec(a).vec(b)=1`. If the length of projection vector of the vector `vec(b)` on the vector `vec(a) xx vec(c )` is l, then the vlaue of `3l^(2)` is equal to ____

Answer

To solve the problem, we need to follow these steps:

### Step 1: Define the vectors
Let \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{c} = \hat{j} - \hat{k}\). We need to find \(\vec{b}\) such that \(\vec{a} \times \vec{b} = \vec{c}\) and \(\vec{a} \cdot \vec{b} = 1\).

### Step 2: Find \(\vec{b}\)
We can express \(\vec{b}\) in terms of its components:
\[ \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 3: Calculate \(\vec{a} \times \vec{b}\)
Using the determinant formula for the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating the determinant gives:
\[ \vec{a} \times \vec{b} = (1z - 1y)\hat{i} - (1z - 1x)\hat{j} + (1y - 1x)\hat{k} = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} \]
Setting this equal to \(\vec{c} = \hat{j} - \hat{k}\), we have:
\[ (z - y) = 0, \quad -(z - x) = 1, \quad (y - x) = -1 \]

### Step 4: Solve the equations
From \(z - y = 0\), we get \(z = y\).
From \(-(z - x) = 1\), substituting \(z = y\) gives:
\[ -(y - x) = 1 \implies x - y = 1 \implies x = y + 1 \]
From \(y - x = -1\), substituting \(x = y + 1\) gives:
\[ y - (y + 1) = -1 \implies -1 = -1 \text{ (always true)} \]
Thus, we can express \(\vec{b}\) as:
\[ \vec{b} = (y + 1)\hat{i} + y\hat{j} + y\hat{k} \]

### Step 5: Use the dot product condition
Now, we use the condition \(\vec{a} \cdot \vec{b} = 1\):
\[ (1)(y + 1) + (1)(y) + (1)(y) = 1 \implies y + 1 + y + y = 1 \implies 3y + 1 = 1 \implies 3y = 0 \implies y = 0 \]
Thus, \(x = 1\) and \(z = 0\), giving us:
\[ \vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i} \]

### Step 6: Calculate \(\vec{a} \times \vec{c}\)
Now we calculate \(\vec{a} \times \vec{c}\):
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} \]
Calculating this determinant gives:
\[ \vec{a} \times \vec{c} = (1 \cdot (-1) - 1 \cdot 1)\hat{i} - (1 \cdot (-1) - 1 \cdot 0)\hat{j} + (1 \cdot 1 - 1 \cdot 0)\hat{k} = (-2)\hat{i} + (1)\hat{j} + (1)\hat{k} \]

### Step 7: Find the magnitude of \(\vec{a} \times \vec{c}\)
The magnitude of \(\vec{a} \times \vec{c}\) is:
\[ |\vec{a} \times \vec{c}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 8: Calculate the projection length \(l\)
The projection length \(l\) of \(\vec{b}\) on \(\vec{a} \times \vec{c}\) is given by:
\[ l = \frac{|\vec{b} \cdot (\vec{a} \times \vec{c})|}{|\vec{a} \times \vec{c}|} \]
Calculating \(\vec{b} \cdot (\vec{a} \times \vec{c})\):
\[ \vec{b} \cdot (\vec{a} \times \vec{c}) = \hat{i} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = -2 \]
Thus:
\[ l = \frac{|-2|}{\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \]

### Step 9: Calculate \(3l^2\)
Now we calculate \(3l^2\):
\[ l^2 = \left(\frac{2}{\sqrt{6}}\right)^2 = \frac{4}{6} = \frac{2}{3} \]
Thus:
\[ 3l^2 = 3 \cdot \frac{2}{3} = 2 \]

### Final Answer
The value of \(3l^2\) is \(2\).

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