Question

Let vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be such that $$ \mathbf{a} = \hat{i} + 2\hat{j} - \hat{k}, \quad \mathbf{b} = 2\hat{i} - \hat{j} + \hat{k}, \quad \mathbf{c} = \hat{i} + \hat{j} + \hat{k} $$ Then the volume of the parallelepiped formed by these vectors is:

• A. -7
• B. -8
• C. 8
• D. -9

Hint:

The volume of a parallelepiped is found using the scalar triple product: \[ V = | \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) | \] Ensure correct determinant calculation and sign handling.

Answer 1

The Correct Option is A

To find the volume of the parallelepiped formed by vectors a, b, and c, we need to calculate the scalar triple product, defined as:

Volume = a ⋅ (b × c).

Given the vectors:

• a = î + 2ĵ - k̂,
• b = 2î - ĵ + k̂,
• c = î + ĵ + k̂.

First, compute the cross product b × c.

The formula for the cross product of two vectors u and v is:

u × v = (u₂v₃ - u₃v₂)î + (u₃v₁ - u₁v₃)ĵ + (u₁v₂ - u₂v₁)k̂.

Applying this formula:

b × c = (−1×1 − 1×1)î + (1×1 − 2×1)ĵ + (2×1 − (−1)×1)k̂.

This simplifies to:

b × c = (−1 − 1)î + (1 − 2)ĵ + (2 + 1)k̂ = −2î − 1ĵ + 3k̂.

Now, compute the dot product a ⋅ (b × c).

The dot product d ⋅ v is given by:

d ⋅ v = d₁v₁ + d₂v₂ + d₃v₃.

Using the values:

a = î + 2ĵ - k̂, b × c = −2î − 1ĵ + 3k̂.

a ⋅ (b × c) = 1×(−2) + 2×(−1) + (−1)×3,

This calculates to:

−2 − 2 − 3 = −7.

Hence, the volume of the parallelepiped is -7.

Answer 2

The volume \( V \) of the parallelepiped formed by three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) is given by the scalar triple product: \[ V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \] Step 1: Compute \( \mathbf{b} \times \mathbf{c} \) Let: \[ \mathbf{b} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}, \quad \mathbf{c} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \] Then, \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}((-1)(1) - (1)(1)) - \hat{j}((2)(1) - (1)(1)) + \hat{k}((2)(1) - (-1)(1)) \] \[ = \hat{i}(-1 - 1) - \hat{j}(2 - 1) + \hat{k}(2 + 1) = -2\hat{i} - \hat{j} + 3\hat{k} \] Step 2: Compute \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) \[ \mathbf{a} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \Rightarrow \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (1)(-2) + (2)(-1) + (-1)(3) = -2 - 2 - 3 = -7 \]