Medium MCQ +4 / -1 PYQ · JEE Mains 2020

A vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$$ lies in the plane of the vectors, $\overrightarrow b = \widehat i + \widehat j$ and $\overrightarrow c = \widehat i - \widehat j + 4\widehat k$. If $\overrightarrow a$ bisects the angle between $\overrightarrow b$ and $\overrightarrow c$, then:

  1. A $\overrightarrow a .\widehat i + 3 = 0$
  2. B $\overrightarrow a .\widehat k - 4 = 0$ Correct answer
  3. C $\overrightarrow a .\widehat i + 1 = 0$
  4. D $\overrightarrow a .\widehat k + 2 = 0$

Solution

Angle bisector $\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$ <br><br>= $$\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$$ <br><br>$\Rightarrow$ $$\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$ <br><br>comparing with $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$ <br><br>${{2\lambda } \over {3\sqrt 2 }}$ = 2 <br><br>$\Rightarrow$ $\lambda$ = ${3\sqrt 2 }$ <br><br>$\therefore$ $\overrightarrow a = \left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$ <br><br>Then $\overrightarrow a .\widehat k - 4$ <br><br>= 4 - 4 = 0

About this question

Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors

This question is part of PrepWiser's free JEE Main question bank. 169 more solved questions on Vector Algebra are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →