Let $\mathrm{ABC}$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{ck}$ and $\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}, \mathrm{~d}0$…

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Area }=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -7 \ 6 & d & -2 \end{array}ight|=15 \sqrt{2} \ & (-4+7 d) \hat{i}-\hat{j}(-2+42)+\hat{k}(d-12) \ & (7 d-4)^2+(40)^2+(d-12)^2=1800 \ & 50 d^2-80 d-40=0 \ & 5 d^2-8 d-4=0 \ & 5 d^2-10 d-2 d-4 \ & 5 d(d-2)+2(d-2)=0 \ & d=2 	ext { or } d=-\frac{2}{5} \ & \because d>0, d=2 \ & (a+1) \hat{i}+(b+2) \hat{j}+(c-7) \hat{k}=6 \hat{i}+2 \hat{j}-2 \hat{k} \ & a+1=6 \& b+2=2, c-7=-2 \ & a=5 \quad b=0 \quad c=5 \ & |A B|=\sqrt{1+4+49}=\sqrt{54} \ & |B C|=\sqrt{25+25}=\sqrt{50} \ & |A C|=\sqrt{86+4+4}=\sqrt{44} \end{aligned}$ Ans. 54

Let $\mathrm{ABC}$ be a triangle of area $15 \sqrt{2}$ and the vectors…

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