Let `veca = 2i + j+k, vecb = i+ 2j -k and a` unit vector `vecc` be coplanar. If `vecc` is pependicular to `veca`. Then `vecc` is

Options

A. `1/sqrt2(-j+k)`
B. `1/sqrt3(i-j-k)`
C. `1/sqrt5(i-2j)`
D. `1/sqrt3(i-j-k)`

Explanation

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

### Step 1: Define the vectors
We are given:
\[ \vec{a} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \]

### Step 2: Express \(\vec{c}\) in terms of \(\vec{a}\) and \(\vec{b}\)
Since \(\vec{c}\) is coplanar with \(\vec{a}\) and \(\vec{b}\), we can express \(\vec{c}\) as a linear combination of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{c} = x\vec{a} + y\vec{b} \]
Substituting the values of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{c} = x(2\hat{i} + \hat{j} + \hat{k}) + y(\hat{i} + 2\hat{j} - \hat{k}) \]
\[ = (2x + y)\hat{i} + (x + 2y)\hat{j} + (x - y)\hat{k} \]

### Step 3: Use the condition that \(\vec{c}\) is perpendicular to \(\vec{a}\)
For \(\vec{c}\) to be perpendicular to \(\vec{a}\), the dot product \(\vec{a} \cdot \vec{c}\) must equal 0:
\[ \vec{a} \cdot \vec{c} = (2\hat{i} + \hat{j} + \hat{k}) \cdot ((2x + y)\hat{i} + (x + 2y)\hat{j} + (x - y)\hat{k}) = 0 \]
Calculating the dot product:
\[ = 2(2x + y) + 1(x + 2y) + 1(x - y) = 0 \]
\[ = 4x + 2y + x + 2y + x - y = 0 \]
\[ = 6x + 3y = 0 \]
This simplifies to:
\[ 2x + y = 0 \quad \Rightarrow \quad y = -2x \]

### Step 4: Substitute \(y\) back into \(\vec{c}\)
Now substitute \(y = -2x\) into the expression for \(\vec{c}\):
\[ \vec{c} = (2x - 2x)\hat{i} + (x + 2(-2x))\hat{j} + (x - (-2x))\hat{k} \]
\[ = 0\hat{i} + (x - 4x)\hat{j} + (x + 2x)\hat{k} \]
\[ = 0\hat{i} - 3x\hat{j} + 3x\hat{k} \]
\[ = 3x(-\hat{j} + \hat{k}) \]

### Step 5: Find the unit vector \(\vec{c}\)
To make \(\vec{c}\) a unit vector, we need to find the magnitude of \(\vec{c}\):
\[ |\vec{c}| = \sqrt{(3x)^2 + (-3x)^2} = \sqrt{9x^2 + 9x^2} = \sqrt{18x^2} = 3\sqrt{2}|x| \]
Setting the magnitude equal to 1:
\[ 3\sqrt{2}|x| = 1 \quad \Rightarrow \quad |x| = \frac{1}{3\sqrt{2}} \]
Thus, \(x = \frac{1}{3\sqrt{2}}\) or \(x = -\frac{1}{3\sqrt{2}}\).

### Step 6: Substitute \(x\) back to find \(\vec{c}\)
Using \(x = \frac{1}{3\sqrt{2}}\):
\[ \vec{c} = 3\left(\frac{1}{3\sqrt{2}}\right)(-\hat{j} + \hat{k}) = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k}) \]
Thus, the unit vector \(\vec{c}\) is:
\[ \vec{c} = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k}) \]

### Final Answer
\[ \vec{c} = \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k}) \]