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Let $\mathrm{\overrightarrow{a}=\alpha \hat{i}+2 \hat{j}-\hat{k} \: and\: \overrightarrow{b}=-2 \hat{i}+\alpha \hat{j}+\hat{k}}$ , where $\mathrm{\alpha \in R}$ . If the area of the parallelogram whose adjacent sides are represented by the vectors $\mathrm{\overrightarrow{a} \: and\: \overrightarrow{b}\: is \sqrt{15\left(\alpha^{2}+4\right)}}$ , then the value of $2|\vec{\mathrm{a}}|^{2}+(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}})|\vec{\mathrm{b}}|^{2}$ is equal to :
Option: 1
$10$

Option: 2
$7$

Option: 3
$9$

Option: 4
$14$

Answers (1)

best_answer

$\text{Area of parallelogram} =|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|$

$\mathrm{=|\left|\begin{array}{ccc} i & j & k \\ \alpha & 2 & -1 \\ -2 & \alpha & 1 \end{array}\right| |} \\$

$\mathrm{=|(2+\alpha) i-(\alpha-2) j+\left(\alpha^{2}+4\right) {k} \mid} \\$

$\mathrm{=\sqrt{(2+\alpha)^{2}+(\alpha-2)^{2}+\left(\alpha^{2}+4\right)^{2}}}$

Given that this area also equals $\mathrm{\sqrt{15\left(\alpha^{2}+4\right)}} \\$

$\mathrm{\Rightarrow(2+\alpha)^{2}+(\alpha-2)^{2}+\left(\alpha^{2}+4\right)^{2}=15\left(\alpha^{2}+4\right) }\\$

$\mathrm{\Rightarrow 2\left(\alpha^{2}+4\right)+\left(\alpha^{2}+4\right)^{2}=15\left(\alpha^{2}+4\right)} \\$

$\mathrm{\Rightarrow\left(\alpha^{2}+4\right)=13 }\\$

$\mathrm{\Rightarrow \alpha^{2}=9}$

$\text{Now for}\: 2|\vec{\mathrm{a}}|^{2}+(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}})|\vec{\mathrm{b}}|^{2}$

$|\vec{\mathrm{a}}|^{2} =\alpha^{2}+4+1=9+5=14 \\$

$\vec{\mathrm{a}} \cdot \vec{\mathrm{b}} =-2 \alpha+2 \alpha-1=-1 \\$

$|\vec{\mathrm{b}}|^{2} =4+\alpha^{2}+1=14 \\$

$\therefore 2|\vec{\mathrm{a}}|^{2}+(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}})|\vec{\mathrm{b}}|^{2} \\$

$\mathrm{=2(14)+(-1)(14) }\\$

$\mathrm{=14}$

Hence the correct answer is option 4

Posted by

manish painkra

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