$\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. Vector $\vec{c}$ is such that $\vec{a}.\vec{c}=4$ and $\vec{a} imes\vec{c}=\vec{b}$, then $\vec{c}=$:

Question

$\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. Vector $\vec{c}$ is such that $\vec{a}.\vec{c}=4$ and $\vec{a}	imes\vec{c}=\vec{b}$, then $\vec{c}=$:

Accepted Answer

To determine why Option 1 is the correct answer for the vector $\vec{c}$, we need to verify that it satisfies both conditions given in the problem: $\vec{a} \cdot \vec{c} = 4$ and $\vec{a} \times \vec{c} = \vec{b}$.

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

**Condition 1: $\vec{a} \cdot \vec{c} = 4$**

Let's assume $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.

The dot product $\vec{a} \cdot \vec{c}$ is calculated as:
$$
\vec{a} \cdot \vec{c} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 2x + y - z
$$

We need:
$$
2x + y - z = 4
$$

**Condition 2: $\vec{a} \times \vec{c} = \vec{b}$**

The cross product $\vec{a} \times \vec{c}$ is calculated using the determinant:
$$
\vec{a} \times \vec{c} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \
2 & 1 & -1 \
x & y & z
\end{vmatrix}
$$

Calculating the determinant:
$$
\vec{a} \times \vec{c} = \hat{i}(1 \cdot z - (-1) \cdot y) - \hat{j}(2 \cdot z - (-1) \cdot x) + \hat{k}(2 \cdot y - 1 \cdot x)
$$
$$
= \hat{i}(z + y) - \hat{j}(2z + x) + \hat{k}(2y - x)
$$

We need:
$$
\hat{i}(z + y) - \hat{j}(2z + x) + \hat{k}(2y - x) = \hat{j} + \hat{k}
$$

This implies:
1. $z + y = 0$
2. $- (2z + x) = 1$ or $2z + x = -1$
3. $2y - x = 1$

**Verification for Option 1: $\vec{c} = \hat{i} + \hat{j} - \hat{k}$**

Substitute $x = 1$, $y = 1$, $z = -1$ into the conditions:

1. $2x + y - z = 2(1) + 1 - (-1) = 2 + 1 + 1 = 4$ (satisfied)
2. $z + y = -1 + 1 = 0$ (satisfied)
3. $2z + x = 2(-1) + 1 = -2 + 1 = -1$ (satisfied)
4. $2y - x = 2(1) - 1 = 2 - 1 = 1$ (satisfied)

Thus, Option 1 satisfies both conditions.

**Why Other Options are Incorrect:**

- **Option 2: $3\hat{i} + \hat{j} + \hat{k}$**
  - $2x + y - z = 2(3) + 1 - 1 = 6 + 1 - 1 = 6 \neq 4$
  - Does not satisfy the dot product condition.

- **Option 3: $\hat{i} + 3\hat{j} + \hat{k}$**
  - $2x + y - z = 2(1) + 3 - 1 = 2 + 3 - 1 = 4$ (satisfied)
  - $z + y = 1 + 3 = 4 \neq 0$
  - Does not satisfy the cross product condition.

- **Option 4: $2\hat{i} - \hat{j} - \hat{k}$**
  - $2x + y - z = 2(2) - 1 - (-1) = 4 - 1 + 1 = 4$ (satisfied)
  - $z + y = -1 - 1 = -2 \neq 0$
  - Does not satisfy the cross product condition.

Therefore, Option 1 is the only choice that satisfies both conditions, confirming it as the correct answer.