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

Let a→,b→ and c→ are lie in the XY plane Let a→=i^,b→=−12i^+32j^ and c→=−12i^−32j^Therefore |p+q+r| =|λa+μb+νc|=|λi^+μ(−12i^+32j^ )+ν(−12i^−32j^)|=|(λ−μ2−ν2)i^+32(μ−ν)j^|=λ2+μ2+ν2−λμ−λν−μν=12(λ−μ)2+(μ−ν)2+(ν−λ)2≥121+1+4=3⇒ |p+q+r| can take a value equal to 3 and 2

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

Let a→,b→ and c→ are lie in the XY plane Let a→=i^,b→=−12i^+32j^ and c→=−12i^−32j^Therefore |p+q+r| =|λa+μb+νc|=|λi^+μ(−12i^+32j^ )+ν(−12i^−32j^)|=|(λ−μ2−ν2)i^+32(μ−ν)j^|=λ2+μ2+ν2−λμ−λν−μν=12(λ−μ)2+(μ−ν)2+(ν−λ)2≥121+1+4=3⇒ |p+q+r| can take a value equal to 3 and 2

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$\vec{a}, \vec{b}$ and $\vec{c}$ are three coplanar unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ . 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 along the following options $| \vec{p} + \vec{q} + \vec{r} |$ can take a value equal to

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$\sqrt{3}$

answer is C.

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Detailed Solution

Let $\vec{a}, \vec{b}$ and $\vec{c}$ are lie in the XY plane
Let $\vec{a} = \hat{i}, \vec{b} = - \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} a n d \vec{c} = - \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}$
Therefore $| p + q + r | = | λ a + μ b + ν c |$
$\begin{array}{l} = | λ \hat{i} + μ (- \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) + ν (- \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) | \\ = | (λ - \frac{μ}{2} - \frac{ν}{2}) \hat{i} + \frac{\sqrt{3}}{2} (μ - ν) \hat{j} | \\ = \sqrt{λ^{2} + μ^{2} + ν^{2} - λ μ - λ ν - μ ν} \end{array}$

$\begin{array}{l} = \frac{1}{\sqrt{2}} \sqrt{{(λ - μ)}^{2} + {(μ - ν)}^{2} + {(ν - λ)}^{2}} \\ \geq \frac{1}{\sqrt{2}} \sqrt{1 + 1 + 4} = \sqrt{3} \end{array}$