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MathsJEE Main 28 Jan 2026, Shift 1·Q25·Numerical·Tough

Vectors · Angle bisector in a triangle

Let PQRPQR be a triangle such that PQ→=−2i^−j^+2k^\overrightarrow{PQ} = -2\hat i - \hat j + 2\hat k and PR→=ai^+bj^−4k^\overrightarrow{PR} = a\hat i + b\hat j - 4\hat k, a,b∈Za, b \in \mathbb{Z}. Let SS be the point on QRQR, which is equidistant from the lines PQPQ and PRPR. If ∣PR→∣=9\lvert\overrightarrow{PR}\rvert = 9 and PS→\overrightarrow{PS} is parallel to i^−7j^+2k^\hat i - 7\hat j + 2\hat k, then the value of 3a−4b3a - 4b is .
We corrected a printing error in the original paper

Source printed “PS→=i^−7j^+2k^\overrightarrow{PS} = \hat i - 7\hat j + 2\hat k”; corrected to the intended “PS→\overrightarrow{PS} is parallel to i^−7j^+2k^\hat i - 7\hat j + 2\hat k”: the printed vector has length 363\sqrt6 and puts SS off the line QRQR (no such triangle; the source marks the question Bonus), while its direction is the internal bisector for a=7a = 7, b=−4b = -4, which the key 37 uses.

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See the full 28 Jan 2026, Shift 1 paper