7.4 Q-1

Question Statement

Compute the cross product $\underline{a} \times \underline{b}$ and $\underline{b} \times \underline{a}$, and check your answer by showing that both $\underline{a}$ and $\underline{b}$ are perpendicular to $\underline{a} \times \underline{b}$ and $\underline{b} \times \underline{a}$.

Given:

• $\underline{\mathbf{a}} = 2 \underline{\mathbf{i}} + \underline{\mathbf{j}} - \underline{\mathbf{k}}$
• $\underline{\mathbf{b}} = \underline{\mathbf{i}} - \underline{\mathbf{j}} + \underline{\mathbf{k}}$

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Background and Explanation

In vector algebra, the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ results in a vector that is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. To compute the cross product $\mathbf{a} \times \mathbf{b}$, we use the determinant of a matrix formed by the unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$ and the components of $\mathbf{a}$ and $\mathbf{b}$:

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{vmatrix}$$

Additionally, to verify if two vectors are perpendicular, we check if their dot product is zero. If the dot product of two vectors $\mathbf{u}$ and $\mathbf{v}$ is zero, i.e. $\mathbf{u} \cdot \mathbf{v} = 0$, then the vectors are perpendicular.

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Solution

Part (i): Compute $\underline{a} \times \underline{b}$

We are given the vectors:

$$\underline{\mathbf{a}} = 2 \underline{\mathbf{i}} + \underline{\mathbf{j}} - \underline{\mathbf{k}}, \quad \underline{\mathbf{b}} = \underline{\mathbf{i}} - \underline{\mathbf{j}} + \underline{\mathbf{k}}$$

We compute the cross product using the determinant formula:

$$\underline{a} \times \underline{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} 2 & 1 & -1 1 & -1 & 1 \end{vmatrix}$$

Expanding the determinant:

$$\underline{a} \times \underline{b} = \mathbf{i}(1 - (-1)) - \mathbf{j}(2 + 1) + \mathbf{k}(-2 - 1)$$ $$\underline{a} \times \underline{b} = 2\mathbf{i} - 3\mathbf{j} - 3\mathbf{k}$$

So, the cross product is:

$$\underline{a} \times \underline{b} = 2\mathbf{i} - 3\mathbf{j} - 3\mathbf{k}$$

Part (ii): Verify Perpendicularity of $\underline{a}$ and $\underline{b}$ with $\underline{a} \times \underline{b}$

First, check if $\underline{a}$ is perpendicular to $\underline{a} \times \underline{b}$ by computing the dot product:

$$\underline{a} \cdot (\underline{a} \times \underline{b}) = (2\mathbf{i} + \mathbf{j} - \mathbf{k}) \cdot (2\mathbf{i} - 3\mathbf{j} - 3\mathbf{k})$$ $$= 2(2) + 1(-3) + (-1)(-3) = 4 - 3 + 3 = 0$$

Since the dot product is zero, $\underline{a} \perp \underline{a} \times \underline{b}$.

Next, check if $\underline{b}$ is perpendicular to $\underline{a} \times \underline{b}$:

$$\underline{b} \cdot (\underline{a} \times \underline{b}) = (\mathbf{i} - \mathbf{j} + \mathbf{k}) \cdot (2\mathbf{i} - 3\mathbf{j} - 3\mathbf{k})$$ $$= 1(2) + (-1)(-3) + 1(-3) = 2 + 3 - 3 = 0$$

Since the dot product is zero, $\underline{b} \perp \underline{a} \times \underline{b}$.

Part (iii): Compute $\underline{b} \times \underline{a}$

Next, we compute the cross product $\underline{b} \times \underline{a}$:

$$\underline{b} \times \underline{a} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} 1 & -1 & 1 2 & 1 & -1 \end{vmatrix}$$

Expanding the determinant:

$$\underline{b} \times \underline{a} = \mathbf{i}(1 - (-1)) - \mathbf{j}(-1 - 2) + \mathbf{k}(1 + 2)$$ $$\underline{b} \times \underline{a} = 0\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$$

So:

$$\underline{b} \times \underline{a} = 3\mathbf{j} + 3\mathbf{k}$$

Part (iv): Verify Perpendicularity of $\underline{b} \times \underline{a}$

Check if $\underline{b} \times \underline{a}$ is perpendicular to $\underline{a}$:

$$(\underline{b} \times \underline{a}) \cdot \underline{a} = (0\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) \cdot (2\mathbf{i} + \mathbf{j} - \mathbf{k})$$ $$= 0(2) + 3(1) + 3(-1) = 0 + 3 - 3 = 0$$

Since the dot product is zero, $\underline{b} \times \underline{a} \perp \underline{a}$.

Check if $\underline{b} \times \underline{a}$ is perpendicular to $\underline{b}$:

$$(\underline{b} \times \underline{a}) \cdot \underline{b} = (0\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) \cdot (\mathbf{i} - \mathbf{j} + \mathbf{k})$$ $$= 0(1) + 3(-1) + 3(1) = 0 - 3 + 3 = 0$$

Since the dot product is zero, $\underline{b} \times \underline{a} \perp \underline{b}$.

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Key Formulas or Methods Used

1. Cross Product:

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{vmatrix}$$

2. Perpendicular Check (Dot Product): Two vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular if:

$$\mathbf{a} \cdot \mathbf{b} = 0$$

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Summary of Steps

1. Compute $\underline{a} \times \underline{b}$ using the determinant formula.
2. Verify if $\underline{a}$ is perpendicular to $\underline{a} \times \underline{b}$ by checking the dot product.
3. Verify if $\underline{b}$ is perpendicular to $\underline{a} \times \underline{b}$ by checking the dot product.
4. Compute $\underline{b} \times \underline{a}$ using the determinant formula.
5. Verify if $\underline{b} \times \underline{a}$ is perpendicular to $\underline{a}$ and $\underline{b}$ by checking the dot product.

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