Question

If the position vectors of the vertices $A , B , C$ of a triangle ABC are $ 7 \hat{j}+10 \hat{k}, -\hat{i}+6 \hat{j}+6 \hat{k}$ and $-4 \hat{i}+9 \hat{j}+6 \hat{k}$ respectively, the triangle is :

• A. equilateral
• B. isosceles
• C. scalene
• D. right angled and isosceles also

Answer 1

Answer: (D)

Given, $\overrightarrow{ OA }=7 \hat{j}+10 \hat{k}, \overrightarrow{ OB }=-\hat{i}+6 \hat{j}+6 \hat{k}$,
$\overrightarrow{ OC }=-4 \hat{i}+9 \hat{j}+6 \hat{k}$
$ \Rightarrow \overrightarrow{ AB }=\overrightarrow{ OB }-\overrightarrow{ OA }=-\hat{i}-\hat{j}-4 \hat{k}$
$\overrightarrow{ BC }=\overrightarrow{ OC } \, \overrightarrow{ OB }=3 \hat{i}|3 \hat{j} ; \overrightarrow{ CA }=\overrightarrow{ OA } \overrightarrow{ OC }=4 \hat{i} 2 \hat{j}+|4 \hat{k}$
$\Rightarrow |\overrightarrow{ AB }|=\sqrt{1^{2}+1^{2}+4^{2}}=\sqrt{18}=3 \sqrt{2}$;
$\Rightarrow |\overrightarrow{ BC }|=\sqrt{3^{2}+3^{2}}=\sqrt{18}=3 \sqrt{2}$;
$\Rightarrow |\overrightarrow{ CA }|=\sqrt{4^{2}+2^{2}+4^{2}}=\sqrt{36}=6$
$3 \sqrt{2}, 3 \sqrt{2}$ & $6$ are sides of a right angled $\Delta$
$\because(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=36=6^{2}$
Hence, the $\Delta\, ABC$ is a right -angled and isosceles also.