Let $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$. Then $(\vec{a}

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Let $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$.

Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to :

$-4$

$4$

$-2$

$2$

The correct answer is

$2$

Vector Calculation: $(\vec{a} - \vec{b}) \cdot \vec{d}$

Given the vectors:

$\vec{a} = -\hat{i} + \hat{j} + 2\hat{k} = \langle -1, 1, 2 \rangle$
$\vec{b} = \hat{i} - \hat{j} - 3\hat{k} = \langle 1, -1, -3 \rangle$

Calculate $\vec{c} = \vec{a} \times \vec{b}$

The cross product $\vec{c}$ is computed using the determinant:

$ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 1 & -1 & -3 \end{vmatrix} $

Expanding the determinant yields:

$ \vec{c} = \hat{i}((1)(-3) - (2)(-1)) - \hat{j}((-1)(-3) - (2)(1)) + \hat{k}((-1)(-1) - (1)(1)) $ $ \vec{c} = \hat{i}(-3 + 2) - \hat{j}(3 - 2) + \hat{k}(1 - 1) $ $ \vec{c} = -\hat{i} - \hat{j} = \langle -1, -1, 0 \rangle $

Calculate $\vec{d} = \vec{c} \times \vec{a}$

The cross product $\vec{d}$ is calculated using $\vec{c}$ and $\vec{a}$:

$ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 0 \\ -1 & 1 & 2 \end{vmatrix} $

Expanding the determinant:

$ \vec{d} = \hat{i}((-1)(2) - (0)(1)) - \hat{j}((-1)(2) - (0)(-1)) + \hat{k}((-1)(1) - (-1)(-1)) $ $ \vec{d} = \hat{i}(-2) - \hat{j}(-2) + \hat{k}(-1 - 1) $ $ \vec{d} = -2\hat{i} + 2\hat{j} - 2\hat{k} = \langle -2, 2, -2 \rangle $

Calculate Difference Vector $\vec{a} - \vec{b}$

Subtract vector $\vec{b}$ from vector $\vec{a}$:

$ \vec{a} - \vec{b} = \langle -1, 1, 2 \rangle - \langle 1, -1, -3 \rangle $ $ \vec{a} - \vec{b} = \langle (-1 - 1), (1 - (-1)), (2 - (-3)) \rangle $ $ \vec{a} - \vec{b} = \langle -2, 2, 5 \rangle $

Calculate Final Dot Product $(\vec{a} - \vec{b}) \cdot \vec{d}$

Compute the dot product of the resulting vectors $(\vec{a} - \vec{b})$ and $\vec{d}$:

$ (\vec{a} - \vec{b}) \cdot \vec{d} = \langle -2, 2, 5 \rangle \cdot \langle -2, 2, -2 \rangle $ $ (\vec{a} - \vec{b}) \cdot \vec{d} = (-2)(-2) + (2)(2) + (5)(-2) $ $ (\vec{a} - \vec{b}) \cdot \vec{d} = 4 + 4 - 10 $ $ (\vec{a} - \vec{b}) \cdot \vec{d} = 8 - 10 = -2 $

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