For some real number $\lambda$, if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$, then the number of distinct possible values of $\lambda$ is

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Aicharya · Accepted Answer

$\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ Area of the triangle $=\frac{\sqrt{195}}{2}$ $\Rightarrow \frac{1}{2}|\vec{a} 	imes \vec{b}|=\frac{1}{2} \sqrt{195} \Rightarrow|\vec{a} 	imes \vec{b}|^2=195$ ...(i) Now, $\vec{a} 	imes \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 3 & -1 & \lambda \ \lambda & 1 & -3\end{array}ight|$ $\begin{aligned} & =\hat{i}(3-\lambda)-\hat{j}\left(-9-\lambda^2ight)+\hat{k}(3+\lambda) \ & |\vec{a} 	imes \vec{b}|=\sqrt{(3-\lambda)^2+\left(9+\lambda^2ight)^2+(3+\lambda)^2}\end{aligned}$ $\Rightarrow|\vec{a} 	imes \vec{b}|^2=(3-\lambda)^2+\left(9+\lambda^2ight)^2+(3+\lambda)^2$ ...(ii) From eqns. (i) and (ii), we get $\begin{aligned} & (3-\lambda)^2+\left(9+\lambda^2ight)^2+(3+\lambda)^2=195 \ & \Rightarrow 9+\lambda^2-6 \lambda+81+\lambda^4+18 \lambda^2+9+\lambda^2+6 \lambda=195 \ & \Rightarrow \lambda^4+20 \lambda^2+99=195 \ & \Rightarrow \lambda^4+20 \lambda^2-96=0 \ & \Rightarrow\left(\lambda^2+24ight)\left(\lambda^2-4ight)=0 \ & \Rightarrow \lambda^2-4=0 	ext { or } \lambda^2+24=0 \ & \Rightarrow \lambda= \pm 2 \Rightarrow \lambda^2=-24 	ext { (not possible) } \ & 	herefore \lambda= \pm 2 .\end{aligned}$

For some real number $\lambda$, if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda…

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