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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Vector Algebra

Let ABC be a triangle of area 15 2 and the vectors AB = i +2 j -7 k...

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{ck} and AC→=6i^+dj^−2k^, d>0\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 _______

Correct answer

0

Step-by-step solution

 Area =12∣i^j^k^12−76d−2∣=152(−4+7d)i^−j^(−2+42)+k^(d−12)(7d−4)2+(40)2+(d−12)2=180050d2−80d−40=05d2−8d−4=05d2−10d−2d−45d(d−2)+2(d−2)=0d=2 or  d=−25∵d>0,d=2(a+1)i^+(b+2)j^+(c−7)k^=6i^+2j^−2k^a+1=6b+2=2,c−7=−2a=5b=0c=5∣AB∣=1+4+49=54∣BC∣=25+25=50∣AC∣=86+4+4=44\begin{aligned} & \text{ Area }=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{array}\right|=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 \text{\text{ 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
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