Question

If the three vectors \( \vec{a} = \hat{i} + \lambda\hat{j} + \hat{k} \), \( \vec{b} = \hat{j} + \hat{k} \), and \( \vec{c} = \hat{i} + \hat{j} \) are coplanar, find the exact value of the scalar constant \( \lambda \).

• A. \( 0 \)
• B. \( 1 \)
• C. \( -2 \)
• D. \( 2 \)

Hint:

The Scalar Triple Product measures the volume of a parallelopiped spanned by three vectors. When vectors are coplanar, the volume collapses to zero, which is why equating the determinant to 0 works perfectly.

Answer 1

The Correct Option is D

Concept: Three non-zero vectors are coplanar (meaning they lie in the exact same geometric plane) if and only if their Scalar Triple Product evaluates to exactly zero: \[ \left[ \vec{a} \quad \vec{b} \quad \vec{c} \right] = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \] This scalar product can be computed efficiently by setting the vector components into a \( 3 \times 3 \) matrix row layout and evaluating its determinant.

Step 1: Set up the matrix determinant using the vector components.
Extract the directional coefficients from vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) to form the rows of the matrix: \[ \left| \begin{matrix} 1 & \lambda & 1 0 & 1 & 1 1 & 1 & 0 \end{matrix} \right| = 0 \]

Step 2: Expand the determinant along the first row.
Evaluate the 2x2 sub-determinants across the top row: \[ 1 \cdot \left| \begin{matrix} 1 & 1 1 & 0 \end{matrix} \right| - \lambda \cdot \left| \begin{matrix} 0 & 1 1 & 0 \end{matrix} \right| + 1 \cdot \left| \begin{matrix} 0 & 1 1 & 1 \end{matrix} \right| = 0 \] Simplify the cross-multiplication pairs: \[ 1(0 - 1) - \lambda(0 - 1) + 1(0 - 1) = 0 \] \[ -1 + \lambda - 1 = 0 \]

Step 3: Isolate the unknown variable \( \lambda \).
Combine the constant numerical values: \[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]