Question

Arrange the following in ascending order of magnitude
$A$: Area of triangle with sides $2\hat{i}-7\hat{j}+\hat{k},4\hat{j}-3\hat{k}$
$B$: Area of triangle with vertices $3\hat{i}+\hat{j},5\hat{i}+2\hat{j}+\hat{k},\hat{i}-2\hat{j}+3\hat{k}$
$C$: Area of triangle with vertices $(1,2,3)$,$(2,5,-1)$ , $(-1,1,2)$
$D$: Area of triangle with sides $\overrightarrow{a}+2\overrightarrow{b}$ and $2\overrightarrow{a}+\overrightarrow{b}$ where $|\overrightarrow{a}|=3,|\overrightarrow{b}|=2,(\overrightarrow{a},\overrightarrow{b})=\frac{\pi }{3}$

• A. $B,A,C,D$
• B. $D,A,C,B$
• C. $D,C,BA$
• D. $B,C,D,A$

Answer 1

Answer: (D)

$B,C,D,A$

Area of a triangle $\Delta ABC=\frac{|AB\times AC|}{2}$
So,
$A)$ Area $=\frac{1}{2}|(2\overline{i}-7\overline{j}+\overline{k})\times (4\overline{j}-3\overline{k})|=\frac{1}{2}|8\overline{k}+6\overline{j}+21\overline{i}-4\overline{i}|$
$=\frac{1}{2}|17\overline{i}+6\overline{j}+8\overline{k}|=\frac{\sqrt{389}}{2}$
$B)$ Sides are $2\overline{i}+\overline{j}+\overline{k}$ and $-2\overline{i}-3\overline{j}+3\overline{k}$
Area $=\frac{1}{2}|(2\overline{i}+\overline{j}+\overline{k})\times (-2\overline{i}-3\overline{j}+3\overline{k})|=\frac{1}{2}|6\overline{i}-8\overline{j}-4\overline{k}|=\frac{\sqrt{116}}{2}$
$C)$ Sides are $2\overline{i}+\overline{j}+\overline{k}$ and $3\overline{i}+4\overline{j}-3\overline{k}$
Area $=\frac{1}{2}|(2\overline{i}+\overline{j}+\overline{k})\times (3\overline{i}+4\overline{j}-3\overline{k})|=\frac{1}{2}|-7\overline{i}+9\overline{j}+5\overline{k}|$
$=\frac{\sqrt{155}}{2}$
$D)$ Area $=\frac{1}{2}|(\overline{a}+2\overline{b})\times (2\overline{a}+\overline{b})|$
$=\frac{1}{2}|3\overline{a}\times \overline{b}|=\frac{3\times 3\times 2\times \sin {60}^o}{2}$
$=\frac{\sqrt{243}}{2}$
Now comparing, we have the increasing order of Areas as $B,C,D,A$.