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QJEEM24Ma1500: JEE Main 2024 Physics - Class 11 JEEM24Ma1500 Solution: Let be a triangle of area and the vectors and . Then the square …

Let ABC\mathrm{ABC} be a triangle of area 15215 \sqrt{2} and the vectors AB→=i^+2j^−7k^,BC→=ai^+bj^+ck^\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k} 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 ABC\mathrm{ABC} is _________.

Step-by-Step Verified Solution

Area of triangle ABC=152A B C=15 \sqrt{2}

amp;⇒12∣AB‾×AC‾∣=152.... (i)amp;AB‾×AC‾∣i^amp;j^amp;k^1amp;2amp;−76amp;damp;−2∣amp;=(7d−4)i^−40j^+(d−12)k^... (ii)\begin{aligned}& \Rightarrow \frac{1}{2}|\overline{A B} \times \overline{A C}|=15 \sqrt{2} \quad \text{.... (i)}\\& \quad \overline{A B} \times \overline{A C}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & -7 \\6 & d & -2\end{array}\right| \\& =(7 d-4) \hat{i}-40 \hat{j}+(d-12) \hat{k} \quad \text{... (ii)}\end{aligned}

From (i) and (ii) 5d2−8d−4=05 d^2-8 d-4=0

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

Also, AB‾+BC‾=AC‾\overline{A B}+\overline{B C}=\overline{A C}

amp;⇒a+1=6⇒a=5amp;b+2=d⇒b=0\begin{aligned}& \Rightarrow a+1=6 \Rightarrow a=5 \\& b+2=d \Rightarrow b=0\end{aligned}

and c−7=−2⇒c=5c-7=-2 \Rightarrow c=5

∣AB‾∣=54,∣AC‾∣=44,∣BC‾∣=50|\overline{A B}|=\sqrt{54},|\overline{A C}|=\sqrt{44},|\overline{B C}|=\sqrt{50}

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

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