Question

Let \[ \vec{a}=2\hat{i}-\hat{j}+\hat{k}, \qquad \vec{b}=\hat{i}+2\hat{j}-\hat{k}, \] and \[ \vec{c}=\lambda\hat{i}+\mu\hat{j}+3\hat{k}. \] If \[ [\vec{a}\ \vec{b}\ \vec{c}]=0 \] and \[ \vec{c}\cdot(\vec{a}+\vec{b})=10, \] then find the value of \(\lambda+\mu\).

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Whenever scalar triple product becomes zero, immediately think of coplanarity and determinant expansion.

Answer 1

The Correct Option is A

Concept: If scalar triple product of three vectors is zero, then the vectors are coplanar. This gives a determinant equation involving unknown parameters.

Step 1: Using scalar triple product condition.
Given: \[ [\vec{a}\ \vec{b}\ \vec{c}]=0 \] Thus, \[ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -1 \lambda & \mu & 3 \end{vmatrix}=0 \] Expanding along first row: \[ 2\begin{vmatrix} 2 & -1 \\ \mu & 3 \end{vmatrix} +1\begin{vmatrix} 1 & -1 \\ \lambda & 3 \end{vmatrix} +1\begin{vmatrix} 1 & 2 \\ \lambda & \mu \end{vmatrix}=0 \] \[ 2(6+\mu)+(3+\lambda)+(\mu-2\lambda)=0 \] \[ 12+2\mu+3+\lambda+\mu-2\lambda=0 \] \[ 15+3\mu-\lambda=0 \] \[ \lambda=3\mu+15 \]

Step 2: Using dot product condition.
Now, \[ \vec{a}+\vec{b} = (2+1)\hat{i}+(-1+2)\hat{j}+(1-1)\hat{k} \] \[ \vec{a}+\vec{b}=3\hat{i}+\hat{j} \] Given: \[ \vec{c}\cdot(\vec{a}+\vec{b})=10 \] Thus, \[ (\lambda\hat{i}+\mu\hat{j}+3\hat{k}) \cdot (3\hat{i}+\hat{j}) =10 \] \[ 3\lambda+\mu=10 \] Substituting \[ \lambda=3\mu+15 \] \[ 3(3\mu+15)+\mu=10 \] \[ 9\mu+45+\mu=10 \] \[ 10\mu=-35 \] \[ \mu=-\frac{7}{2} \] Then, \[ \lambda=3\left(-\frac72\right)+15 \] \[ \lambda=-\frac{21}{2}+\frac{30}{2} =\frac92 \] Therefore, \[ \lambda+\mu = \frac92-\frac72 = 1 \] Hence the required answer is: \[ \boxed{(A)\ 1} \]