Q46 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
Let $P Q R$ be a triangle such that $\overrightarrow{P Q}=-2 \hat{i}-\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{PR}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}}-4 \hat{\mathrm{k}}, a, b \in \mathbb{Z}$. Let S be the point on QR, which is equidistant from the lines PQ and PR. If $|\overrightarrow{\mathrm{PR}}|=9$ and $\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then the value of $3 a-4 b$ is $\_\_\_\_$
Correct answer: 37
Solution
Given: $\vec{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$, $\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$ ($a, b \in \mathbb{Z}$), $\vec{PS} = \hat{i} - 7\hat{j} + 2\hat{k}$, $|\vec{PR}| = 9$
From $|\vec{PR}| = 9$: $a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65$ ...(1)
Since S is on QR equidistant from lines PQ and PR, PS is the angle bisector. So:
$\cos\theta = \frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}||\vec{PS}|} = \frac{-2+7+4}{3 \cdot 3\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}}$
Similarly: $\frac{1}{\sqrt{6}} = \frac{\vec{PS} \cdot \vec{PR}}{|\vec{PS}||\vec{PR}|} = \frac{a - 7b - 8}{3\sqrt{6} \cdot 9}$
$a - 7b = 35$ ...(2)
From (1) and (2): $a = 7, b = -4$
$3a - 4b = 21 + 16 = 37$ (NTA Answer)
However, $\vec{QS} = 3\hat{i} - 6\hat{j}$ and $\vec{SR} = 6\hat{i} + 3\hat{j} - 6\hat{k}$ are not parallel, so Q, S, R are not collinear. Also $\cos\theta = \frac{2\sqrt{2}}{\sqrt{3}} > 1$, which is impossible. Hence no such triangle exists.
So, this should be a bonus question.
From $|\vec{PR}| = 9$: $a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65$ ...(1)
Since S is on QR equidistant from lines PQ and PR, PS is the angle bisector. So:
$\cos\theta = \frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}||\vec{PS}|} = \frac{-2+7+4}{3 \cdot 3\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}}$
Similarly: $\frac{1}{\sqrt{6}} = \frac{\vec{PS} \cdot \vec{PR}}{|\vec{PS}||\vec{PR}|} = \frac{a - 7b - 8}{3\sqrt{6} \cdot 9}$
$a - 7b = 35$ ...(2)
From (1) and (2): $a = 7, b = -4$
$3a - 4b = 21 + 16 = 37$ (NTA Answer)
However, $\vec{QS} = 3\hat{i} - 6\hat{j}$ and $\vec{SR} = 6\hat{i} + 3\hat{j} - 6\hat{k}$ are not parallel, so Q, S, R are not collinear. Also $\cos\theta = \frac{2\sqrt{2}}{\sqrt{3}} > 1$, which is impossible. Hence no such triangle exists.
So, this should be a bonus question.
Hence, the correct answer is 37.