First slide

Algebra of vectors

Question

$\vec{a}, \vec{b} and \vec{c}$ are three coplanar unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0 .$ If three vectors $\vec{p}, \vec{q} and \vec{r}$ are parallel to $\vec{a}, \vec{b} and \vec{c}$ respectively, and have integral but different magnitudes, then among the following options $| \vec{p} + \vec{q} + \vec{r} |$ can take a value equal to

Moderate

A

1
B

0
C

$\sqrt{3}$
D

2

Solution

$Let \vec{a}, \vec{b} and \vec{c} lie in the x - y plane.$
$Let \vec{a} = \hat{i}, \vec{b} = - \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{i} and \vec{c} = - \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}$
Therefore,
$\begin{array}{l} | \vec{p} + \vec{q} + \vec{r} | = | λ \vec{a} + μ \vec{b} + v \vec{c} | \\ = |λ \hat{i} + μ (- \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) + v (- \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})| \\ = |(λ - \frac{μ}{2} - \frac{v}{2}) \hat{i} + \frac{\sqrt{3}}{2} (μ - v) \hat{j}| \\ = \sqrt{{(λ - \frac{μ}{2} - \frac{v}{2})}^{2} + \frac{3}{4} (μ - v)^{2}} \end{array}$
$\begin{array}{l} = \sqrt{λ^{2} + μ^{2} + v^{2} - λμ - λv - μv} \\ = \frac{1}{\sqrt{2}} \sqrt{(λ - μ)^{2} + (μ - v)^{2} + (v - λ)^{2}} \\ \geq \frac{1}{\sqrt{2}} \sqrt{1 + 1 + 4} = \sqrt{3} \end{array}$
Hence $| \vec{p} + \vec{q} + \vec{r} |$ can take a value equal to $\sqrt{3}$ and 2

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