If the points P, Q, R with position vectors $5\hat i +λ\hat j, 20\hat i-\hat j$ and $15\hat i-6\hat j$ respectively are collinear, then the value of $λ$ is

-16

The correct answer is Option (1) → -16

Let position vectors be:

$\vec{P} = 5\hat{i} + \lambda\hat{j}$

$\vec{Q} = 20\hat{i} - \hat{j}$

$\vec{R} = 15\hat{i} - 6\hat{j}$

Points P, Q, R are collinear ⟺ vectors $\vec{PQ}$ and $\vec{PR}$ are linearly dependent ⟺ $\vec{PQ} \times \vec{PR} = \vec{0}$

$\vec{PQ} = \vec{Q} - \vec{P} = (20 - 5)\hat{i} + (-1 - \lambda)\hat{j} = 15\hat{i} - (\lambda + 1)\hat{j}$

$\vec{PR} = \vec{R} - \vec{P} = (15 - 5)\hat{i} + (-6 - \lambda)\hat{j} = 10\hat{i} - (\lambda + 6)\hat{j}$

Now compute $\vec{PQ} \times \vec{PR}$:

$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 15 & -(\lambda + 1) & 0 \\ 10 & -(\lambda + 6) & 0 \end{vmatrix}$

Only $\hat{k}$ component survives:

$= \hat{k} \left(15(-\lambda - 6) - 10(-\lambda - 1)\right) = \hat{k}(-15\lambda - 90 + 10\lambda + 10)$

$= \hat{k}(-5\lambda - 80)$

Set this equal to zero for collinearity:

$-5\lambda - 80 = 0 \Rightarrow \lambda = -16$