Let ABC be a triangle of area 15√2 and the vectors AB=i+2j-7k, BC=ai+bj+ck and AC=6i+dj-2k…

Question

Let $ABC$ be a triangle of area $15\sqrt{2}$ and the vectors $\overset{	o \leftarrow }{AB}=\hat{i}+2\hat{j}-7\hat{k},\overset{	o \leftarrow }{BC}=a\hat{i}+b\hat{j}+c\hat{k}$ and $\overset{	o \leftarrow }{AC}=6\hat{i}+d\hat{j}-2\hat{k},d>0$. Then the square of the length of the largest side of the triangle $ABC$ is \_\_\_\_\_\_\_\_\_.

Answer: 54

Accepted Answer

Area of triangle $ABC=15\sqrt{2}$

$\begin{gathered} \Rightarrow \frac{1}{2}|\vec{AB}	imes \vec{AC}|=15\sqrt{2}\, \ \, \vec{AB}	imes \vec{AC}|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  2 \quad  -7 \ 6 \quad  d \quad  -2\end{array}| \ =(7d-4)\hat{i}-40\hat{j}+(d-12)\hat{k}\, \end{gathered}$

From (i) and (ii) $5{d}^{2}-8d-4=0$

$\Rightarrow d=\frac{-}{5}$ (Rejected) or $d=2$

Also, $\vec{AB}+\vec{BC}=\vec{AC}$

$\begin{gathered} \Rightarrow a+1=6\Rightarrow a=5 \ b+2=d\Rightarrow b=0 \end{gathered}$

and $c-7=-2\Rightarrow c=5$

$|\vec{AB}|=\sqrt{54},|\vec{AC}|=\sqrt{44},|\vec{BC}|=\sqrt{50}$

Largest side has length of $\sqrt{54}$ units

Solutions · 1