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QJEEM24Mabb0d: JEE Main 2024 Physics - Class 11 JEEM24Mabb0d Solution: The position vectors of the vertices and of a triangle are and…

The position vectors of the vertices A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} of a triangle are 2i^−3j^+3k^,2i^+2j^+3k^2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k} and −i^+j^+3k^-\hat{i}+\hat{j}+3 \hat{k} respectively. Let ll denotes the length of the angle bisector AD\mathrm{AD} of ∠BAC\angle \mathrm{BAC} where D\mathrm{D} is on the line segment BC\mathrm{BC} , then 2l22 l^2 equals :

Multiple Choice Options

  • A 45
  • B 50
  • C 42
  • D 49

Step-by-Step Verified Solution

amp;AB=5amp;AC=5\begin{aligned}& \mathrm{AB}=5 \\& \mathrm{AC}=5\end{aligned}

JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Vector Algebra Question 45 English Explanation

∴D\therefore \mathrm{D} is midpoint of BC\mathrm{BC}

amp;D(12,32,3)amp;∴l=(2−12)2+(−3−32)2+(3−3)2amp;l=452amp;∴2l2=45\begin{aligned}& \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \\& \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \\& l=\sqrt{\frac{45}{2}} \\& \therefore 2 l^2=45\end{aligned}

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