Let $\vec{a}=2\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c}=2\hat{i}-3\hat{j}+4\hat{k}.$ A vector $\vec{r}$ satisfying $\vec{r} imes\vec{b}=\vec{c} imes\vec{b}$ and $\vec{r}.\vec{a}=0$ is:

Question

Let $\vec{a}=2\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c}=2\hat{i}-3\hat{j}+4\hat{k}.$ A vector $\vec{r}$ satisfying $\vec{r}	imes\vec{b}=\vec{c}	imes\vec{b}$ and $\vec{r}.\vec{a}=0$ is:

Accepted Answer

To solve the problem, we need to find a vector $\vec{r}$ that satisfies the conditions $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$.

### Step 1: Calculate $\vec{c} \times \vec{b}$

Given:
$$
\vec{b} = \hat{i} + 2\hat{j} + \hat{k}
$$
$$
\vec{c} = 2\hat{i} - 3\hat{j} + 4\hat{k}
$$

The cross product $\vec{c} \times \vec{b}$ is calculated as follows:
$$
\vec{c} \times \vec{b} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \
2 & -3 & 4 \
1 & 2 & 1
\end{vmatrix}
$$

Expanding the determinant:
$$
= \hat{i}((-3)(1) - (4)(2)) - \hat{j}((2)(1) - (4)(1)) + \hat{k}((2)(2) - (-3)(1))
$$
$$
= \hat{i}(-3 - 8) - \hat{j}(2 - 4) + \hat{k}(4 + 3)
$$
$$
= \hat{i}(-11) + \hat{j}(2) + \hat{k}(7)
$$
$$
= -11\hat{i} + 2\hat{j} + 7\hat{k}
$$

### Step 2: Find $\vec{r}$ such that $\vec{r} \times \vec{b} = -11\hat{i} + 2\hat{j} + 7\hat{k}$

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then:
$$
\vec{r} \times \vec{b} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \
x & y & z \
1 & 2 & 1
\end{vmatrix}
$$

Expanding the determinant:
$$
= \hat{i}(y \cdot 1 - z \cdot 2) - \hat{j}(x \cdot 1 - z \cdot 1) + \hat{k}(x \cdot 2 - y \cdot 1)
$$
$$
= \hat{i}(y - 2z) - \hat{j}(x - z) + \hat{k}(2x - y)
$$

Equating to $-11\hat{i} + 2\hat{j} + 7\hat{k}$:
$$
y - 2z = -11
$$
$$
-(x - z) = 2 \quad \Rightarrow \quad x - z = -2
$$
$$
2x - y = 7
$$

### Step 3: Solve the system of equations

From $x - z = -2$, we have:
$$ x = z - 2 $$

Substitute $x = z - 2$ into $2x - y = 7$:
$$
2(z - 2) - y = 7
$$
$$
2z - 4 - y = 7
$$
$$
2z - y = 11
$$

Now solve the system:
1. $y - 2z = -11$
2. $2z - y = 11$

Add the two equations:
$$
(y - 2z) + (2z - y) = -11 + 11
$$
$$
0 = 0
$$

This confirms the system is consistent. Solve for $y$ and $z$:
From $y - 2z = -11$:
$$ y = 2z - 11 $$

Substitute into $2z - y = 11$:
$$
2z - (2z - 11) = 11
$$
$$
11 = 11
$$

This confirms the equations are consistent. Choose $z = 3$ (arbitrary choice to satisfy the system):
$$ y = 2(3) - 11 = 6 - 11 = -5 $$
$$ x = z - 2 = 3 - 2 = 1 $$

Thus, $\vec{r} = \hat{i} - 5\hat{j} + 3\hat{k}$.

### Step 4: Verify $\vec{r} \cdot \vec{a} = 0$

Given $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$:
$$
\vec{r} \cdot \vec{a} = (1)(2) + (-5)(1) + (3)(1)
$$
$$
= 2 - 5 + 3 = 0
$$

Thus, $\vec{r} \cdot \vec{a} = 0$ is satisfied.

### Conclusion

The vector $\vec{r} = \hat{i} - 5\hat{j} + 3\hat{k}$ satisfies both conditions, confirming that Option 4 is correct. The other options do not satisfy both conditions, as they do not yield the correct cross product or dot product results.