Verified Solution • 2025–26 Session moderate

QJEEM16Mab0f7: JEE Main 2016 Physics - Class 11 JEEM16Mab0f7 Solution: In a triangle ABC, right angled at the vertex A, if the…

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3 i^\widehat i + j^\widehat j −- k^\widehat k ,    −- i^\widehat i + 3 j^\widehat j + p k^\widehat k and 5 i^\widehat i + q j^\widehat j −- 4 k^\widehat k , then the point (p, q) lies on a line :

Multiple Choice Options

  • A parallel to x-axis.
  • B parallel to y-axis.
  • C making an acute angle with the positive direction of x-axis.
  • D making an obtuse angle with the positive direction of x-axis.

Step-by-Step Verified Solution

Given,

A→=3i^+j^−k^\overrightarrow A = 3\widehat i + \widehat j - \widehat k

B→=−i^+3j^−pk^\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k

C→=5i^+9j^−4k^\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k

∴\therefore     AB→=−4i^+2j^+(p+1)k^\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k

       AC→=2i^+(q−1)j^−3k^\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k

JEE Main 2016 (Online) 9th April Morning Slot Mathematics - Vector Algebra Question 198 English Explanation

Δ\Delta ABC is a right angle triangle.

Here AB→\overrightarrow {AB} perpendicular to AC→\overrightarrow {AC}

∴\therefore     AB→\overrightarrow {AB}   .   AC→\overrightarrow {AC}   =  0

⇒\Rightarrow     −- 8 + 2(q −- 1) −- 3(p + 1) = 0

⇒\Rightarrow    3p −- 2q + 13 = 0

∴\therefore    (p, q) lies on the line

3x −- 2y + 13 = 0

And slope of the line = 32{3 \over 2}

∴\therefore    line makes an angle less than 90o or acute angle with the positive direction of x-axis.

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring