Question

The volume of the parallelepiped whose co-terminous edges are $\hat{i} + \hat{j}, \hat{i} + \hat{k}, \hat{i} + \hat{j}$ is:

• A. $6$ cu. units
• B. $2$ cu. units
• C. $4$ cu. units
• D. $3$ cu. units

Answer 1

The Correct Option is B

1. Identify the edge vectors:

The given edges are \( \vec{a} = \hat{i} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{k} \), and \( \vec{c} = \hat{i} + \hat{j} \).

Note: There seems to be a typo in the problem statement (two identical vectors). Assuming the correct edges are \( \hat{i} + \hat{k} \), \( \hat{i} + \hat{j} \), and \( \hat{i} + \hat{k} \).

2. Compute the scalar triple product:

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = 1(1 \cdot 1 - 0 \cdot 0) - 0(1 \cdot 1 - 0 \cdot 1) + 1(1 \cdot 0 - 1 \cdot 1) = 1 - 0 - 1 = 0 \]

This gives zero volume, which suggests a different interpretation. Alternatively, if the edges are distinct (e.g., \( \hat{i} + \hat{k} \), \( \hat{i} + \hat{j} \), \( \hat{j} + \hat{k} \)):

\[ \begin{vmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 1(1 \cdot 1 - 0 \cdot 1) - 0(1 \cdot 1 - 0 \cdot 0) + 1(1 \cdot 1 - 1 \cdot 0) = 1 + 0 + 1 = 2 \]

Correct Answer: (B) 2 cu. units

Answer 2

The volume of a parallelepiped is given by the scalar triple product: \[ V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \] Where: - $\vec{a} = \hat{i} + \hat{j}$ - $\vec{b} = \hat{i} + \hat{k}$ - $\vec{c} = \hat{i} + \hat{j}$ First, compute the cross product $\vec{b} \times \vec{c}$: \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(-1) - \hat{j}(1) + \hat{k}(1) = -\hat{i} - \hat{j} + \hat{k} \] Next, compute the dot product $\vec{a} \cdot (\vec{b} \times \vec{c})$: \[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (1)(-1) + (1)(-1) + (0)(1) = -2 \] Thus, the volume is: \[ V = | -2 | = 2 \, \text{cu. units} \] Hence, the correct answer is $2$ cu. units.

Answer 3

The volume of a parallelepiped formed by three vectors $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ is given by the absolute value of their scalar triple product: $ |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| $.

The co-terminous edges are given as:

• $ \mathbf{a} = \hat{j} + \hat{k} = \langle 0, 1, 1 \rangle $
• $ \mathbf{b} = \hat{i} + \hat{k} = \langle 1, 0, 1 \rangle $
• $ \mathbf{c} = \hat{i} + \hat{j} = \langle 1, 1, 0 \rangle $

We compute the scalar triple product:

$$ \mathbf{b} \times \mathbf{c} = \langle 1, 0, 1 \rangle \times \langle 1, 1, 0 \rangle = \langle (0 \cdot 0 - 1 \cdot 1), (1 \cdot 1 - 1 \cdot 0), (1 \cdot 1 - 0 \cdot 1) \rangle = \langle -1, 1, 1 \rangle $$ $$ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \langle 0, 1, 1 \rangle \cdot \langle -1, 1, 1 \rangle = (0)(-1) + (1)(1) + (1)(1) = 2 $$

Therefore, the volume of the parallelepiped is $ |2| = 2 $ cubic units.