Let vectors V1¯=(cosθ)i^−(3sinθ)j^−2k^, V2¯=(2cosθ)i^−(sinθ)j^+k^ and V3¯=(cosθ)i^−(2sinθ)j^+3k^. Represent sides of a triangle, where θ∈[0,π2). If area of the triangle is minimum and Length of the median which bisects the side represented by V3¯ is l, then find the value of 1024500 (2l2) .

V¯1×V¯2=|i^j^k^cosθ−3sinθ−22cosθ−sinθ1|=−5sinθ i^ −5cosθj^+5sinθ cosθk^ ∴ Area of the Δ=12|V¯1×V¯2|12.51+sin2θcos2θ=521+14sin22θ For area of Δ to be minimum sin2θ=0 ⇒θ=0 Now, V¯1=i^−2k^,V¯2=2i^+k^,V¯3=i^+3k^|V¯1|=5,|V¯2|=5,|V¯3|=10I =102⇒2I2=5

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Let vectors $\bar{V_{1}} = (\cos θ) \hat{i} - (3 \sin θ) \hat{j} - 2 \hat{k},$ $\bar{V_{2}} = (2 \cos θ) \hat{i} - (\sin θ) \hat{j} + \hat{k}$ and $\bar{V_{3}} = (\cos θ) \hat{i} - (2 \sin θ) \hat{j} + 3 \hat{k} .$ Represent sides of a triangle, where $θ \in [0, \frac{π}{2})$ . If area of the triangle is minimum and Length of the median which bisects the side represented by $\bar{V_{3}}$ is $l,$ then find the value of $\frac{1024}{500} (2 l^{2})$ .

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Detailed Solution

$\begin{array}{l} {\bar{V}}_{1} \times {\bar{V}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \cos θ & - 3 \sin θ & - 2 \\ 2 \cos θ & - \sin θ & 1 \end{matrix} | \\ = - 5 \sin θ \hat{i} - 5 \cos θ \hat{j} + 5 \sin θ \cos θ \hat{k} \end{array}$
$∴$ Area of the $Δ = \frac{1}{2} | {\bar{V}}_{1} \times {\bar{V}}_{2} |$
$\frac{1}{2} .5 \sqrt{1 + \sin^{2} θ \cos^{2} θ} = \frac{5}{2} \sqrt{1 + \frac{1}{4} \sin^{2} 2 θ}$
For area of $Δ$ to be minimum $\sin 2 θ = 0 \Rightarrow θ = 0$
Now, ${\bar{V}}_{1} = \hat{i} - 2 \hat{k}, {\bar{V}}_{2} = 2 \hat{i} + \hat{k}, {\bar{V}}_{3} = \hat{i} + 3 \hat{k}$
$\begin{array}{l} | {\bar{V}}_{1} | = \sqrt{5}, | {\bar{V}}_{2} | = \sqrt{5}, | {\bar{V}}_{3} | = \sqrt{10} \\ I = \frac{\sqrt{10}}{2} \Rightarrow 2 I^{2} = 5 \end{array}$