Question

Let \(\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}\), \(\vec{b} = \hat{i} + \lambda\hat{j} + 4\hat{k}\) and \(\vec{c} = 2\hat{i} + 4\hat{j} + (\lambda^2 - 1)\hat{k}\) be coplanar vectors. Then the non-zero vector \(\vec{a} \times \vec{c}\) is :

• A. \(-10\hat{i} - 5\hat{j}\)
• B. \(-10\hat{i} + 5\hat{j}\)
• C. \(-14\hat{i} + 5\hat{j}\)
• D. \(-14\hat{i} - 5\hat{j}\)

Hint:

Coplanarity means the scalar triple product vanishes; find the admissible values of lambda, then compute \(\vec a\times\vec c\) and keep the non-zero one.

Answer 1

The Correct Option is B

Coplanar: \(\begin{vmatrix}1&2&4\\1&\lambda&4\\2&4&\lambda^2-1\end{vmatrix}=0\).
Expanding: \(1[\lambda(\lambda^2-1)-16] - 2[(\lambda^2-1)-8] + 4[4-2\lambda] = \lambda^3 - 2\lambda^2 - 9\lambda + 18 = (\lambda-2)(\lambda^2-9)\). So \(\lambda = 2, 3, -3\).
\(\vec a\times\vec c = \left(2(\lambda^2-1)-16\right)\hat i - \left((\lambda^2-1)-8\right)\hat j + (4-4)\hat k = (2\lambda^2-18)\hat i + (9-\lambda^2)\hat j\).
For \(\lambda=\pm3\) this is the zero vector. For \(\lambda=2\): \(-10\hat i + 5\hat j\).
Answer: (2) \(-10\hat{i} + 5\hat{j}\)