Question

Let the position vectors of points \(A\) and \(B\) be \(\hat{i}+\hat{j}+\hat{k}\) and \(2\hat{i}+\hat{j}+3\hat{k}\), respectively. A point \(P\) divides the line segment \(AB\) internally in the ratio \(\lambda : 1\) (\(\lambda \gt 0\)). If \(O\) is the origin and
\(\overrightarrow{OB}\cdot\overrightarrow{OP}-3\left|\overrightarrow{OA}\times\overrightarrow{OP}\right|^{2}=6\), then \(\lambda\) is equal to _______.

Hint:

Write \(\vec{OP}=\frac{\lambda\vec{OB}+\vec{OA}}{\lambda+1}\) and compute the dot and cross products.

Answer 1

Correct Answer: 0.8

\(\vec{OP}=\frac{\lambda\vec{OB}+\vec{OA}}{\lambda+1}=\frac{(2\lambda+1)\hat i+(\lambda+1)\hat j+(3\lambda+1)\hat k}{\lambda+1}\).
\(\vec{OB}\cdot\vec{OP}=\frac{2(2\lambda+1)+(\lambda+1)+3(3\lambda+1)}{\lambda+1}=\frac{14\lambda+6}{\lambda+1}\).
\(\vec{OA}\times\vec{OP}=\frac{1}{\lambda+1}\left(2\lambda\,\hat i-\lambda\,\hat j-\lambda\,\hat k\right)\), so \(\left|\vec{OA}\times\vec{OP}\right|^2=\frac{6\lambda^2}{(\lambda+1)^2}\).
Equation: \(\frac{14\lambda+6}{\lambda+1}-\frac{18\lambda^2}{(\lambda+1)^2}=6\).
\((14\lambda+6)(\lambda+1)-18\lambda^2=6(\lambda+1)^2\Rightarrow -4\lambda^2+20\lambda+6=6\lambda^2+12\lambda+6\).
\(10\lambda^2-8\lambda=0\Rightarrow \lambda=\frac45=0.8\) (since \(\lambda\gt0\)).
Answer: 0.8