Let `a=2hati+hatj+hatk,b=hati+2hatj-hatk` and c is a unit vector coplanar to them. If c is perpendicular to a, then c is equal to

Options

A. `(1)/(sqrt(2))(-hatj+hatk)`
B. `-(1)/(sqrt(3))(hati+hatj+hatk)`
C. `(1)/(sqrt(5))(hati-2hatj)`
D. `(1)/(sqrt(3))(hati+hatj+hatk)`

Answer

To solve the problem, we need to find the unit vector \( \mathbf{c} \) that is coplanar with the vectors \( \mathbf{a} \) and \( \mathbf{b} \), and is also perpendicular to \( \mathbf{a} \).

Given:
- \( \mathbf{a} = 2\hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k} \)

### Step 1: Express \( \mathbf{c} \) as a linear combination of \( \mathbf{a} \) and \( \mathbf{b} \)

Since \( \mathbf{c} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \), we can express \( \mathbf{c} \) as:
\[ \mathbf{c} = x\mathbf{a} + y\mathbf{b} \]
where \( x \) and \( y \) are scalars.

### Step 2: Substitute \( \mathbf{a} \) and \( \mathbf{b} \)

Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{c} = x(2\hat{i} + \hat{j} + \hat{k}) + y(\hat{i} + 2\hat{j} - \hat{k}) \]
Expanding this gives:
\[ \mathbf{c} = (2x + y)\hat{i} + (x + 2y)\hat{j} + (x - y)\hat{k} \]

### Step 3: Use the condition that \( \mathbf{c} \) is perpendicular to \( \mathbf{a} \)

For \( \mathbf{c} \) to be perpendicular to \( \mathbf{a} \), their dot product must equal zero:
\[ \mathbf{a} \cdot \mathbf{c} = 0 \]
Calculating the dot product:
\[ (2\hat{i} + \hat{j} + \hat{k}) \cdot ((2x + y)\hat{i} + (x + 2y)\hat{j} + (x - y)\hat{k}) = 0 \]
This results in:
\[ 2(2x + y) + 1(x + 2y) + 1(x - y) = 0 \]
Simplifying this gives:
\[ 4x + 2y + x + 2y + x - y = 0 \]
Combining like terms:
\[ 6x + 3y = 0 \]
From this, we can express \( y \) in terms of \( x \):
\[ y = -2x \]

### Step 4: Substitute \( y \) back into \( \mathbf{c} \)

Now substituting \( y = -2x \) back into the expression for \( \mathbf{c} \):
\[ \mathbf{c} = (2x - 2x)\hat{i} + (x - 4x)\hat{j} + (x + 2x)\hat{k} \]
This simplifies to:
\[ \mathbf{c} = 0\hat{i} - 3x\hat{j} + 3x\hat{k} \]
Thus, we have:
\[ \mathbf{c} = -3x\hat{j} + 3x\hat{k} \]

### Step 5: Normalize \( \mathbf{c} \) to make it a unit vector

To make \( \mathbf{c} \) a unit vector, we need to set its magnitude to 1:
\[ |\mathbf{c}| = \sqrt{(0)^2 + (-3x)^2 + (3x)^2} = \sqrt{9x^2 + 9x^2} = \sqrt{18x^2} = 3\sqrt{2}|x| \]
Setting this equal to 1:
\[ 3\sqrt{2}|x| = 1 \implies |x| = \frac{1}{3\sqrt{2}} \]

### Step 6: Determine \( \mathbf{c} \)

Now substituting \( x = \frac{1}{3\sqrt{2}} \) into \( \mathbf{c} \):
\[ \mathbf{c} = -3\left(\frac{1}{3\sqrt{2}}\right)\hat{j} + 3\left(\frac{1}{3\sqrt{2}}\right)\hat{k} = -\frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} \]
Thus, the unit vector \( \mathbf{c} \) is:
\[ \mathbf{c} = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k}) \]

### Final Answer

\[ \mathbf{c} = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k}) \]