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MHTCET 23 Apr 2025 Shift 1 Paper

Question : 102 of 150

Marks: +1, -0

Let

\overline{a}, \overline{b}, \overline{c}, \overline{d}

\overline{a} \times \overline{b}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}

\overline{c} \times \overline{d}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}

\left.\begin{array}{cc}\overline{a} \cdot \overline{c} & \overline{b} \cdot \overline{c} \\ \overline{a} \cdot \overline{d} & \overline{b} \cdot \overline{d}\end{array} \right\rvert\,=0

\lambda=

Solution:

Concept:

Use Lagrange's identity:

(\overline{a} \times \overline{b}) \cdot (\overline{c} \times \overline{d}) = (\overline{a} \cdot \overline{c})(\overline{b} \cdot \overline{d}) - (\overline{a} \cdot \overline{d})(\overline{b} \cdot \overline{c})

Explanation:

The given determinant is:

\left|\begin{array}{cc} \overline{a} \cdot \overline{c} & \overline{b} \cdot \overline{c} \\ \overline{a} \cdot \overline{d} & \overline{b} \cdot \overline{d} \end{array}\right| = 0

Expanding this determinant yields:

(\overline{a} \cdot \overline{c})(\overline{b} \cdot \overline{d}) - (\overline{a} \cdot \overline{d})(\overline{b} \cdot \overline{c}) = 0

By Lagrange's identity, the left-hand side is exactly

(\overline{a} \times \overline{b}) \cdot (\overline{c} \times \overline{d})

Therefore, we obtain:

(\overline{a} \times \overline{b}) \cdot (\overline{c} \times \overline{d}) = 0

Substitute the given cross product values:

(2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} - \hat{\mathrm{k}}) \cdot (3\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \lambda\hat{\mathrm{k}}) = 0

Compute the dot product term by term:

2(3) + 3(2) + (-1)(\lambda) = 0

6 + 6 - \lambda = 0

12 - \lambda = 0

\lambda = 12

Answer:

\lambda = 12

, which matches option C.

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