Let three vectors $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of the triangle is $5 \sqrt{6}$.…

Question

Let three vectors $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number, then $|\vec{c}|^2$ is equal to:

Aicharya · Accepted Answer

$\begin{aligned} & \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \\ & \Rightarrow(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\alpha-5,1,-2) \\ & \Rightarrow \mathrm{x}=\alpha-5, \mathrm{y}=1, \mathrm{z}=-2.....(1)\end{aligned}$

Area of $\Delta=5 \sqrt{6}$ (given) $\begin{aligned} & \frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=5 \sqrt{6} \\ & \left\|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \alpha & 4 & 2 \\ \mathrm{x} & 1 & -2\end{array}\right\|=10 \sqrt{6}\end{aligned}$ $\begin{aligned} & \Rightarrow|-10 \hat{\mathrm{i}}-\hat{\mathrm{j}}(-2 \alpha-2 \mathrm{x})+\hat{\mathrm{k}}(\alpha-4 \mathrm{x})|=10 \sqrt{6} \\ & \Rightarrow(2 \alpha+2 \alpha-10)^2+(\alpha-4 \alpha+20)^2=500 \\ & \Rightarrow(4 \alpha-10)^2+(20-3 \alpha)^2=500 \\ & \Rightarrow 25 \alpha^2-80 \alpha-120 \alpha=0 \\ & \Rightarrow \alpha(25 \alpha-200)=0 \\ & \Rightarrow \alpha=8 \text { (given } \alpha \text { is +ve number) } \\ & \Rightarrow \mathrm{x}=\alpha-5=3 \\ & |\overrightarrow{\mathrm{c}}|^2=\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 \\ & =9+1+4 \\ & =14\end{aligned}$

Let three vectors $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k},…

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