Let `vec(a)=2hati-hatj+hatk,vec(b)=hati+2hatj-hatk` and `vec( c )=hati+hatj-2hatk` be three vectors. A vector in the plane of `vec(b)` and `vec( c )` whose projection on `vec(a)` is of magnitude `sqrt(2//3)` is

Options

A. `2hati+3hatj-3vec(b)`
B. `2hati+3hatj+3hatk`
C. `-2hati-hatj+5hatk`
D. `2hati+hatj+5hatk`

Answer

To solve the problem, we need to find a vector \( \vec{r} \) that lies in the plane formed by vectors \( \vec{b} \) and \( \vec{c} \), and whose projection onto vector \( \vec{a} \) has a magnitude of \( \sqrt{\frac{2}{3}} \).

### Step 1: Define the vectors
Given:
- \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \)
- \( \vec{c} = \hat{i} + \hat{j} - 2\hat{k} \)

### Step 2: Express vector \( \vec{r} \)
Since \( \vec{r} \) lies in the plane of \( \vec{b} \) and \( \vec{c} \), we can express \( \vec{r} \) as a linear combination of \( \vec{b} \) and \( \vec{c} \):
\[ \vec{r} = \alpha \vec{b} + \beta \vec{c} \]
Substituting the expressions for \( \vec{b} \) and \( \vec{c} \):
\[ \vec{r} = \alpha (\hat{i} + 2\hat{j} - \hat{k}) + \beta (\hat{i} + \hat{j} - 2\hat{k}) \]
\[ \vec{r} = (\alpha + \beta)\hat{i} + (2\alpha + \beta)\hat{j} + (-\alpha - 2\beta)\hat{k} \]

### Step 3: Calculate the projection of \( \vec{r} \) onto \( \vec{a} \)
The projection of \( \vec{r} \) onto \( \vec{a} \) is given by:
\[ \text{proj}_{\vec{a}} \vec{r} = \frac{\vec{r} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} \]
First, we need to compute \( \vec{r} \cdot \vec{a} \):
\[ \vec{r} \cdot \vec{a} = (\alpha + \beta)(2) + (2\alpha + \beta)(-1) + (-\alpha - 2\beta)(1) \]
Expanding this:
\[ = 2(\alpha + \beta) - (2\alpha + \beta) - (\alpha + 2\beta) \]
\[ = 2\alpha + 2\beta - 2\alpha - \beta - \alpha - 2\beta \]
\[ = -\beta - \alpha \]

### Step 4: Calculate the magnitude of \( \vec{a} \)
Next, we calculate \( |\vec{a}| \):
\[ |\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]
Thus, \( |\vec{a}|^2 = 6 \).

### Step 5: Set up the equation for the magnitude of the projection
We know the magnitude of the projection is \( \sqrt{\frac{2}{3}} \):
\[ \left| \frac{\vec{r} \cdot \vec{a}}{|\vec{a}|} \right| = \sqrt{\frac{2}{3}} \]
Substituting the values:
\[ \left| \frac{-\beta - \alpha}{\sqrt{6}} \right| = \sqrt{\frac{2}{3}} \]
Squaring both sides:
\[ \frac{(-\beta - \alpha)^2}{6} = \frac{2}{3} \]
Multiplying through by 6:
\[ (-\beta - \alpha)^2 = 4 \]
Taking the square root:
\[ -\beta - \alpha = 2 \quad \text{or} \quad -\beta - \alpha = -2 \]
This gives us two equations:
1. \( \beta + \alpha = -2 \)
2. \( \beta + \alpha = 2 \)

### Step 6: Solve for \( \beta \) in terms of \( \alpha \)
From the first equation:
\[ \beta = -2 - \alpha \]
Substituting this into the expression for \( \vec{r} \):
\[ \vec{r} = \alpha \vec{b} + (-2 - \alpha) \vec{c} \]

### Step 7: Substitute and simplify
Substituting \( \beta \) into the expression for \( \vec{r} \):
\[ \vec{r} = \alpha(\hat{i} + 2\hat{j} - \hat{k}) + (-2 - \alpha)(\hat{i} + \hat{j} - 2\hat{k}) \]
This simplifies to:
\[ \vec{r} = \alpha \hat{i} + 2\alpha \hat{j} - \alpha \hat{k} - 2\hat{i} - 2\hat{j} + 4\hat{k} - \alpha \hat{i} - \alpha \hat{j} + 2\alpha \hat{k} \]
Combining like terms gives:
\[ \vec{r} = (-2)\hat{i} + (2\alpha - 2 - \alpha)\hat{j} + (4 - \alpha + 2\alpha)\hat{k} \]
\[ = -2\hat{i} + (\alpha - 2)\hat{j} + (4 + \alpha)\hat{k} \]

### Final Result
The vector \( \vec{r} \) can be expressed in terms of \( \alpha \):
\[ \vec{r} = -2\hat{i} + (\alpha - 2)\hat{j} + (4 + \alpha)\hat{k} \]
You can choose any value for \( \alpha \) to get a specific vector in the plane of \( \vec{b} \) and \( \vec{c} \).