Question

The perimeter of the triangle whose vertices are the points $2\overline{i}-\overline{j}+\overline{k},\overline{i}-3\overline{j}-5\overline{K},3\overline{i}-4\overline{j}-4\overline{j}-4\overline{k}$ is

• A. $\sqrt{3}+\sqrt{35}+\sqrt{7}$
• B. $\sqrt{6}+\sqrt{35}+\sqrt{7}$
• C. $\sqrt{6}+\sqrt{30}+\sqrt{41}$
• D. $\sqrt{6}+\sqrt{35}+\sqrt{41}$

Answer 1

Answer: (D)

Given position vector

$A=2i-j+k$

$B=i-3j+k$

$C=3i-4j-4k$

Perimeter$=|\overline{AB}|+|\overline{BC}|+|\overline{CA}|$

$|\overline{AB}|=\sqrt{(2-1{)}^2+(-1+3{)}^2+(1+5{)}^2}=\sqrt{1+4+36}=\sqrt{41}$

$|\overline{BC}|=\sqrt{(3-1{)}^2+(-4+3{)}^2+(-4+5{)}^2}=\sqrt{4+1+1}=\sqrt{6}$

$|\overline{CA}|=\sqrt{(3-2{)}^2+(-4+1{)}^2+(-4-1{)}^2}=\sqrt{1+9+25}=\sqrt{35}$

$\therefore$ Perimeter$=\sqrt{6}+\sqrt{35}+\sqrt{41}$.