If `veca = hati +hatj , vecb = 2hatj - hatk " and " vecr xx veca = vecb xx vec a , vecr xx vec b = veca xx vec b` , then what is the value of `(vec r)/(|vec r|)`

Options

A. `((hati + 3hatj -hatk))/(sqrt11)`
B. `((hati - 3hatj + hatk))/(sqrt11)`
C. `((hati + 3hatj + hatk))/(sqrt11)`
D. `((hati -3hatj -hatk))/(sqrt11)`

Answer

To solve the problem, we need to find the unit vector of vector **R** given the conditions involving cross products. Let's break down the steps:

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

Let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).

### Step 2: Compute \(\vec{r} \times \vec{a}\)
Using the determinant form for the cross product:

\[ \vec{r} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{r} \times \vec{a} = \hat{i}(y \cdot 0 - z \cdot 1) - \hat{j}(x \cdot 0 - z \cdot 1) + \hat{k}(x \cdot 1 - y \cdot 1) \]
\[ = -z\hat{i} + z\hat{j} + (x - y)\hat{k} \]
\[ = -z\hat{i} + z\hat{j} + (x - y)\hat{k} \]

### Step 3: Compute \(\vec{b} \times \vec{a}\)
Now, we compute \(\vec{b} \times \vec{a}\):

\[ \vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{b} \times \vec{a} = \hat{i}(2 \cdot 0 - (-1) \cdot 1) - \hat{j}(0 \cdot 0 - (-1) \cdot 1) + \hat{k}(0 \cdot 1 - 2 \cdot 1) \]
\[ = \hat{i}(1) - \hat{j}(1) - 2\hat{k} \]
\[ = \hat{i} - \hat{j} - 2\hat{k} \]

### Step 4: Set up the equation
From the problem, we have:

\[ \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \]

This gives us:

\[ -z\hat{i} + z\hat{j} + (x - y)\hat{k} = \hat{i} - \hat{j} - 2\hat{k} \]

### Step 5: Equate coefficients
From the above equation, we can equate the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):

1. For \(\hat{i}\): \(-z = 1\) ⇒ \(z = -1\)
2. For \(\hat{j}\): \(z = -1\) ⇒ \(z = -1\) (consistent)
3. For \(\hat{k}\): \(x - y = -2\)

### Step 6: Solve for \(x\) and \(y\)
Now we need to solve for \(x\) and \(y\) using the second condition:

\[ \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \]

Calculating \(\vec{a} \times \vec{b}\):

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 2 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}(1 \cdot (-1) - 0 \cdot 2) - \hat{j}(1 \cdot (-1) - 0 \cdot 0) + \hat{k}(1 \cdot 2 - 1 \cdot 0) \]
\[ = -\hat{i} + \hat{j} + 2\hat{k} \]

Now we set up the equation:

\[ \vec{r} \times \vec{b} = -\hat{i} + \hat{j} + 2\hat{k} \]

### Step 7: Compute \(\vec{r} \times \vec{b}\)
\[ \vec{r} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 0 & 2 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}(y \cdot (-1) - z \cdot 2) - \hat{j}(x \cdot (-1) - z \cdot 0) + \hat{k}(x \cdot 2 - y \cdot 0) \]
\[ = -y - 2z\hat{i} + x\hat{j} + 2x\hat{k} \]

### Step 8: Equate coefficients again
Setting this equal to \(-\hat{i} + \hat{j} + 2\hat{k}\):

1. For \(\hat{i}\): \(-y - 2z = -1\)
2. For \(\hat{j}\): \(x = 1\)
3. For \(\hat{k}\): \(2x = 2\)

From \(2x = 2\), we find \(x = 1\). Substituting \(z = -1\) into \(-y - 2(-1) = -1\):

\[ -y + 2 = -1 \Rightarrow -y = -3 \Rightarrow y = 3 \]

### Step 9: Final vector \(\vec{r}\)
Thus, we have:

\[ \vec{r} = 1\hat{i} + 3\hat{j} - 1\hat{k} \]

### Step 10: Find the unit vector \(\frac{\vec{r}}{|\vec{r}|}\)
Now, we need to find \(|\vec{r}|\):

\[ |\vec{r}| = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \]

Thus, the unit vector is:

\[ \frac{\vec{r}}{|\vec{r}|} = \frac{1\hat{i} + 3\hat{j} - 1\hat{k}}{\sqrt{11}} = \frac{1}{\sqrt{11}}\hat{i} + \frac{3}{\sqrt{11}}\hat{j} - \frac{1}{\sqrt{11}}\hat{k} \]

### Final Answer
The value of \(\frac{\vec{r}}{|\vec{r}|}\) is:

\[ \frac{1}{\sqrt{11}}\hat{i} + \frac{3}{\sqrt{11}}\hat{j} - \frac{1}{\sqrt{11}}\hat{k} \]