The value of λ for which the vectors 3i - 6j + k and 2i - 4j + λk are…

The correct answer is Option (1) → $\frac{2}{3}$ ##

Since, two vectors are parallel i.e., angle between them is zero.

$∴(3\hat{i} - 6\hat{j} + \hat{k}) \cdot (2\hat{i} - 4\hat{j} + \lambda\hat{k}) = |3\hat{i} - 6\hat{j} + \hat{k}| \cdot |2\hat{i} - 4\hat{j} + \lambda\hat{k}|$

$[∵\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos 0^\circ \Rightarrow \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|]$ $∴\cos 0^\circ = 1)$

$\Rightarrow 6 + 24 + \lambda = \sqrt{9 + 36 + 1} \sqrt{4 + 16 + \lambda^2}$

$\Rightarrow 30 + \lambda = \sqrt{46} \sqrt{20 + \lambda^2}$

$\Rightarrow 900 + \lambda^2 + 60\lambda = 46(20 + \lambda^2) \quad \text{[on squaring both sides]}$

$\Rightarrow \lambda^2 + 60\lambda - 46\lambda^2 = 920 - 900$

$\Rightarrow -45\lambda^2 + 60\lambda - 20 = 0$

$\Rightarrow -45\lambda^2 + 30\lambda + 30\lambda - 20 = 0$

$\Rightarrow -15\lambda(3\lambda - 2) + 10(3\lambda - 2) = 0$

$\Rightarrow (10 - 15\lambda)(3\lambda - 2) = 0$

$∴\lambda = \frac{2}{3}, \frac{2}{3}$