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Let $u$ and $v$ be two vectors in $\mathbf{R}^{2}$ whose Euclidean norms satisfy $\left \| u \right \| = 2\left \| v \right \|$. What is the value of $\alpha$ such that $w = u + \alpha v$ bisects the angle between $u$ and $v$?

  1. $2$
  2. $\frac{1}{2}$
  3. $1$
  4. $\frac{ -1}{2}$
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5 Answers

Best answer
25 votes
25 votes
Angle between $u$ and $w =$ Angle between $w$ and $v$

$\frac{\vec{u}. \vec{w}}{\|u\| \|w\| } = \frac{\vec{w}. \vec{v}}{\|w\| \|v\| }$

$\vec{u}. \vec{w} = 2 \vec{w}. \vec{v}$

$(\alpha -2)\vec{u}. \vec{v} = 2(\alpha -2) \|v\|^2$

$\text{LHS}$ and $\text{RHS}$ would be equal for $\alpha=2$. Hence, correct answer is $(A)$.
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We know that  " The resultant of two equal vectors bisects the angle between them ".

In the given question, $$\overrightarrow{w}  = \overrightarrow{u}  + \alpha \overrightarrow{v} $$

Here, $\overrightarrow{w}$ is the resultant of two vectors $\overrightarrow{u}$ and $\alpha$$\overrightarrow{v}$. 

Thus, for $\overrightarrow{w}$ to bisect the angle between $\overrightarrow{u}$ and $\overrightarrow{v}$, magnitude of the two vectors must be equal.

                                      $\left \| u \right \| =\left \| \alpha v \right \|$               $\Rightarrow$           $2\left \| v \right \| =\alpha \left \| v \right \|$

$\therefore$         $\alpha = 2$

3 votes
3 votes

The geometric shape can also be parallelogram

Answer:

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