Let A=(1,1,1), B=(2,1,3) (position vectors), and P divide AB internally in ratio :1 (0).

Question

Let $A=(1,1,1)$, $B=(2,1,3)$ (position vectors), and $P$ divide $AB$ internally in ratio $\lambda:1$ ($\lambda>0$). If $\vec{OB}\cdot\vec{OP}-3|\vec{OA}	imes\vec{OP}|^2=6$, then $\lambda$ is \_\_\_\_\_\_\_.

Answer: 0.8

Accepted Answer

$$P = \frac{A+\lambda B}{1+\lambda} = \left(\frac{1+2\lambda}{1+\lambda},\ 1,\ \frac{1+3\lambda}{1+\lambda}\right)$$

Substituting into $\vec{OB}\cdot\vec{OP}-3|\vec{OA}\times\vec{OP}|^2=6$ and solving the resulting equation in $\lambda$ (as in the source) gives:

$$\lambda = 0.8$$

Solutions · 1