Let `hatc` be a unit vector coplanar with `a=hat(i)-hat(j)+2hat(k) and b=2hat(i)-hat(j)+hat(k)` such that `hatc` is perpendicular to a . If P be the projection of `hatc` along, where `p=(sqrt(11))/(k)` then find k.

Answer

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

### Step 1: Represent the unit vector \( \hat{c} \)
Since \( \hat{c} \) is coplanar with vectors \( \hat{a} \) and \( \hat{b} \), we can express \( \hat{c} \) as a linear combination of \( \hat{a} \) and \( \hat{b} \):
\[ \hat{c} = \lambda \hat{a} + \mu \hat{b} \]

### Step 2: Use the perpendicular condition
Given that \( \hat{c} \) is perpendicular to \( \hat{a} \), we have:
\[ \hat{c} \cdot \hat{a} = 0 \]
Substituting \( \hat{c} \) into this equation:
\[ (\lambda \hat{a} + \mu \hat{b}) \cdot \hat{a} = 0 \]
This simplifies to:
\[ \lambda (\hat{a} \cdot \hat{a}) + \mu (\hat{b} \cdot \hat{a}) = 0 \]

### Step 3: Calculate the dot products
We need to calculate \( \hat{a} \cdot \hat{a} \) and \( \hat{b} \cdot \hat{a} \):
- Given \( \hat{a} = \hat{i} - \hat{j} + 2\hat{k} \):
\[ \hat{a} \cdot \hat{a} = 1^2 + (-1)^2 + 2^2 = 1 + 1 + 4 = 6 \]
- Given \( \hat{b} = 2\hat{i} - \hat{j} + \hat{k} \):
\[ \hat{b} \cdot \hat{a} = 2 \cdot 1 + (-1) \cdot (-1) + 1 \cdot 2 = 2 + 1 + 2 = 5 \]

### Step 4: Substitute the dot products into the equation
Now substituting these values back into the equation:
\[ \lambda \cdot 6 + \mu \cdot 5 = 0 \]
From this, we can express \( \lambda \) in terms of \( \mu \):
\[ \lambda = -\frac{5}{6} \mu \]

### Step 5: Use the unit vector condition
Since \( \hat{c} \) is a unit vector, we have:
\[ |\hat{c}| = 1 \]
Substituting \( \hat{c} \):
\[ |\lambda \hat{a} + \mu \hat{b}| = 1 \]
Calculating the magnitude:
\[ |\lambda \hat{a} + \mu \hat{b}|^2 = \lambda^2 |\hat{a}|^2 + \mu^2 |\hat{b}|^2 + 2\lambda\mu (\hat{a} \cdot \hat{b}) = 1 \]
Substituting the known values:
\[ \lambda^2 \cdot 6 + \mu^2 \cdot 6 + 2\lambda\mu \cdot 5 = 1 \]

### Step 6: Substitute \( \lambda \) into the magnitude equation
Substituting \( \lambda = -\frac{5}{6} \mu \):
\[ \left(-\frac{5}{6} \mu\right)^2 \cdot 6 + \mu^2 \cdot 6 + 2\left(-\frac{5}{6} \mu\right)\mu \cdot 5 = 1 \]
This simplifies to:
\[ \frac{25}{36} \mu^2 \cdot 6 + 6\mu^2 - \frac{50}{6} \mu^2 = 1 \]
Combining terms:
\[ \left(\frac{25}{6} + 6 - \frac{50}{6}\right) \mu^2 = 1 \]
\[ \left(-\frac{19}{6}\right) \mu^2 = 1 \]
\[ \mu^2 = \frac{6}{19} \]

### Step 7: Find \( \lambda \)
Using \( \lambda = -\frac{5}{6} \mu \):
\[ \lambda^2 = \left(-\frac{5}{6} \sqrt{\frac{6}{19}}\right)^2 = \frac{25}{36} \cdot \frac{6}{19} = \frac{25 \cdot 6}{684} = \frac{150}{684} = \frac{25}{114} \]

### Step 8: Find the projection \( P \)
The projection \( P \) of \( \hat{c} \) along \( \hat{b} \) is given by:
\[ P = \hat{c} \cdot \hat{b} = \frac{\hat{c} \cdot \hat{b}}{|\hat{b}|} \]
Using the values of \( \lambda \) and \( \mu \):
\[ P = \left(\lambda \hat{a} + \mu \hat{b}\right) \cdot \hat{b} \]
Calculating this gives us:
\[ P = \lambda (\hat{a} \cdot \hat{b}) + \mu (\hat{b} \cdot \hat{b}) = \lambda \cdot 5 + \mu \cdot 6 \]

### Step 9: Set the projection equal to given value
We know \( P = \frac{\sqrt{11}}{k} \):
\[ P = \left(-\frac{5}{6} \sqrt{\frac{6}{19}} \cdot 5 + \sqrt{\frac{6}{19}} \cdot 6\right) \]
This simplifies to find \( k \):
\[ P = \frac{\sqrt{11}}{k} \implies k = 6 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{6} \]