Aspire Faculty ID #11625 · Topic: NIMCET 2024 · Just now
NIMCET 2024

The value of m for which volume of the parallelepiped is 4 cubic units whose three edges are represented by a = mi + j + k, b = i – j + k, c = i + 2j –k is

Solution

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

  • $\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
  • $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
  • $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$:

$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k} $

Step 2: Compute dot product with $\vec{a}$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

  • Case 1: $-m + 5 = 4 \Rightarrow m = 1$
  • Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

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