Question

Let \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}\) are vectors such that \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(\overline{\mathrm{c}} \times \overline{\mathrm{d}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\) and if \(\left|\begin{array}{ll}\overline{\mathrm{a}} \cdot \overline{\mathrm{c}} & \overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\ \overline{\mathrm{a}} \cdot \overline{\mathrm{d}} & \overline{\mathrm{b}} \cdot \overline{\mathrm{d}}\end{array}\right|=0\), then \(\lambda=\)

• A. 6
• B. -6
• C. 12
• D. -12

Hint:

The determinant of dot products equals the dot product of the two cross products (Lagrange identity).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The identity \((\bar a\times\bar b)\cdot(\bar c\times\bar d)=(\bar a\cdot\bar c)(\bar b\cdot\bar d)-(\bar b\cdot\bar c)(\bar a\cdot\bar d)\) holds.

Step 2: Apply to the given determinant
The determinant \((\bar a\cdot\bar c)(\bar b\cdot\bar d)-(\bar b\cdot\bar c)(\bar a\cdot\bar d)\) equals \((\bar a\times\bar b)\cdot(\bar c\times\bar d)\), and it is given to be 0.

Step 3: Calculation
\((2\hat i+3\hat j-\hat k)\cdot(3\hat i+2\hat j+\lambda\hat k)=6+6-\lambda=0\)
So \(\lambda=12\).

Final Answer:
\(\lambda=12\). \[ \boxed{\text{(C) }12} \]