Question:medium

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 $\lambda$.

Show 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!
Updated On: May 29, 2026
  • \( 4 \)
  • \( -4 \)
  • \( 2 \)
  • \( -2 \)
Show Solution

The Correct Option is B

Solution and Explanation

To find the value of the scalar parameter \(\lambda\) for which the vectors \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\)\(\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}\), and \(\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}\) are coplanar, we use the condition that the volume of the parallelepiped formed by these vectors is zero.

This condition translates to saying that the scalar triple product of these vectors is zero:

\(V = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0\)

Let's calculate the cross product \(\vec{b} \times \vec{c}\) first:

The cross product \(\vec{b} \times \vec{c}\) is given by the determinant:

\(\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix}\)

Expanding this determinant using the cofactor expansion along the first row:

\(= \hat{i} \left( 2 \cdot 5 - (-3) \cdot \lambda \right) - \hat{j} \left( 1 \cdot 5 - (-3) \cdot 3 \right) + \hat{k} \left( 1 \cdot \lambda - 2 \cdot 3 \right)\)

\(= \hat{i} (10 + 3\lambda) - \hat{j} (5 + 9) + \hat{k} (\lambda - 6)\)

\(= (10 + 3\lambda)\hat{i} - 14\hat{j} + (\lambda - 6)\hat{k}\)

Now compute the dot product \(\vec{a} \cdot (\vec{b} \times \vec{c})\):

\(\vec{a} \cdot (\vec{b} \times \vec{c}) = (2\hat{i} - \hat{j} + \hat{k}) \cdot \left((10 + 3\lambda)\hat{i} - 14\hat{j} + (\lambda - 6)\hat{k}\right)\)

Compute the dot product:

\(= 2(10 + 3\lambda) - 1(-14) + 1(\lambda - 6)\)

\(= 20 + 6\lambda + 14 + \lambda - 6\)

\(= 28 + 7\lambda\)

Set this to zero for the vectors to be coplanar:

\(28 + 7\lambda = 0\)

Solve for \(\lambda\):

\(7\lambda = -28\)

\(\lambda = -\frac{28}{7} = -4\)

Thus, the correct value of the scalar parameter \(\lambda\) is \(-4\).

Was this answer helpful?
0