Mathematics - Vector Question with Solution | TestHub

Let the given vectors be $\overrightarrow{A}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}, \overrightarrow{B}=2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}, \overrightarrow{C}=\left(\mathrm{\alpha }+1\right)\hat{\mathrm{i}}+2\hat{\mathrm{k}} \mathrm{and} \overrightarrow{D}=9\hat{\mathrm{i}}+\left(\mathrm{\alpha }-8\right)\hat{\mathrm{j}}+6\hat{\mathrm{k}}$

We know that if the vectors are coplanar then $\left[\begin{array}{ccc}\overrightarrow{AB}& \overrightarrow{AC}& \overrightarrow{AD}\end{array}\right]=0$

$\Rightarrow \overrightarrow{AB}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)-\left(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)=\hat{i}-\hat{j}+\hat{k}$

$\Rightarrow \overrightarrow{AC}=\left(\left(\mathrm{\alpha }+1\right)\hat{\mathrm{i}}+2\hat{\mathrm{k}}\right)-\left(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)=\alpha \hat{i}+2\hat{j}-\hat{k}$

$\Rightarrow \overrightarrow{AD}=\left(9\hat{\mathrm{i}}+\left(\mathrm{\alpha }-8\right)\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right)-\left(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)=8\hat{i}+\left(\alpha -6\right)\hat{j}+3k$

Now,

$\Rightarrow \left|\begin{array}{ccc}1& -1& 1\\ \alpha & 2& -1\\ 8& \alpha -6& 3\end{array}\right|=0$

$\Rightarrow 1\left(6+\alpha -6\right)+\left(3\alpha +8\right)+\left({\alpha }^2-6\alpha -16\right)=0$

On Simplifying we get,

$\Rightarrow {\alpha }^2-2\alpha -8=0$

Therefore, sum of the roots is $-\left(-2\right)=2$.