Let $ PQR $ be a triangle such that $ \overrightarrow{PQ} = -2\hat{i} - \hat{j} + 2\hat{k} $ and $ \overrightarrow{PR} = a\hat{i} + b\hat{j} - 4\hat{k},; a, b \in \mathbb{Z} $. Let $ S $ be the point on $ QR $, which is equidistant from the lines $ PQ $ and $ PR $. If $ |PR| = 9 $ and $ \overrightarrow{PS} = \hat{i} - 7\hat{j} + 2\hat{k} $, then the value of $ 3a - 4b $ is ______

Solution

$ |PR| = 9 \Rightarrow a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65 $

$ \cos\theta = \frac{\overrightarrow{PQ}\cdot \overrightarrow{PS}}{|PQ||PS|} = \frac{-2 + 7 + 4}{3\cdot 3\sqrt{6}} = \frac{9}{3\cdot 3\sqrt{6}} = \frac{1}{\sqrt{6}} $

Using geometry relations

$ 3a - 4b = 37 $