Question

If the vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$, and $\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}$ represent the concurrent coterminous edges of a parallelopiped whose volume is $0$ (i.e., the vectors are coplanar), find the value of the scalar parameter $\lambda$.

• A. \( 4 \)
• B. \( -4 \)
• C. \( 2 \)
• D. \( -2 \)

Hint:

To cross-verify your answer instantly in the exam hall without expanding the whole determinant again, plug \(\lambda = -4\) directly back into the third row, making it \(\begin{pmatrix} 3 & -4 & 5 \end{pmatrix}\). Notice a neat linear dependency shortcut: if you add the first row \(\begin{pmatrix} 2 & -1 & 1 \end{pmatrix}\) and the second row \(\begin{pmatrix} 1 & 2 -3 \end{pmatrix}\) multiplied by \(-1\), do they show relationships? Checking row choices saves you from silly sign errors!

Answer 1

The Correct Option is B

Concept: The volume (\(V\)) of a parallelopiped formed by three concurrent coterminous vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) is equal to the absolute value of their Scalar Triple Product (STP): \[ V = \left| \vec{a} \cdot (\vec{b} \times \vec{c}) \right| \] When the volume of the parallelopiped is exactly zero, it physically means the structure has flattened completely into a single flat two-dimensional space. Therefore, setting the scalar triple product to zero (\([\vec{a} \;\; \vec{b} \;\; \vec{c}] = 0\)) is the standard mathematical condition to check if three vectors are coplanar. This Scalar Triple Product is evaluated easily by calculating the determinant of the \(3 \times 3\) matrix formed by the components of the three vectors.

Step 1: Setting up the matrix determinant equation.
Since the vectors are coplanar, their determinant must be equal to zero: \[ \begin{vmatrix} 2 & -1 & 1 1 & 2 & -3 3 & \lambda & 5 \end{vmatrix} = 0 \]

Step 2: Expanding the $3 \times 3$ determinant to isolate $\lambda$.
Expanding the determinant along the first row: \[ 2 \cdot \begin{vmatrix} 2 & -3 \lambda & 5 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -3 3 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 3 & \lambda \end{vmatrix} = 0 \] Evaluating the individual \(2 \times 2\) sub-determinants via cross-multiplication: \[ 2 \cdot \Big( (2 \times 5) - (-3 \times \lambda) \Big) + 1 \cdot \Big( (1 \times 5) - (-3 \times 3) \Big) + 1 \cdot \Big( (1 \times \lambda) - (2 \times 3) \Big) = 0 \] Simplifying the terms inside the brackets carefully: \[ 2 \cdot (10 + 3\lambda) + 1 \cdot (5 + 9) + 1 \cdot (\lambda - 6) = 0 \] \[ 20 + 6\lambda + 14 + \lambda - 6 = 0 \]

Step 3: Solving the linear algebraic equation.
Group the coefficients of \(\lambda\) and the constant terms together: \[ (6\lambda + \lambda) + (20 + 14 - 6) = 0 \] \[ 7\lambda + 28 = 0 \] Shift the constant to the right-hand side: \[ 7\lambda = -28 \] \[ \lambda = \frac{-28}{7} = -4 \] This mathematically matches Option (B) perfectly with zero ambiguity.