Question

The value of ${}^{\prime }{a}^{\prime }$ for which the points $A,B,C$ with position vectors $2\overline{i}-\overline{j}+\overline{k},\overline{i}-3\overline{j}-5\overline{k}$ and $a\overline{i}-3\overline{j}+\overline{k}$ are the vertices of a right angled triangle with $\angle C=\frac{\pi }{2}$ are :

• A. $-2$ and $1$
• B. $2$ and $-1$
• C. $2$ and $1$
• D. $-2$ and $-1$

Answer 1

The correct option is C $2$ and $1$

$\angle c=\frac{\pi }{2}$

$\overrightarrow{OA}=2\hat{i}-\hat{j}+\hat{k}\overrightarrow{OB}=\hat{i}-\hat{j}-\hat{k}$

$\overrightarrow{OC}=a\hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}\overrightarrow{OA}=(a-2)\hat{i}\widehat{-}j$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}-\hat{i}-2\hat{j}-6\hat{k}$

$\overrightarrow{BC}=(a-1)\hat{i}6\hat{k}$

$\angle c=\frac{\pi }{2}$

$|AC{|}^2+|BC{|}^2=|AB{|}^2$

$\Rightarrow {\left(\sqrt{(a-2{)}^2+(-2{)}^2}\right)}^2+{\left(\sqrt{(a-1{)}^2+6^2}\right)}^2={\left(\sqrt{1^2+2^2+6^2}\right)}^2$

$\Rightarrow (a-2{)}^2+4+(a-1{)}^2+6^2=1^2+2^2+6^2$

$\Rightarrow 2a^2+4+1-6a=1$

$\Rightarrow a^a-3a+2=0$

$a(a-1)-1(a-2)=0$

$(a-1)(a-2)=0$

$\therefore a=1and2$