A point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and…

A point $C$ with position vector $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}$ (where $\bar{a}, \bar{b}$ and $\bar{c}$ are non coplanar vectors) divides the line joining $A$ and $B$ in the ratio $2: 1$. If the position vector of $A$ is $\bar{a}-2 \bar{b}+3 \bar{c}$, then the position vector of $B$ is
  1. $2 \bar{a}+3 \bar{b}-4 \bar{c}$
  2. $2 \bar{a}-3 \bar{b}+4 \bar{c}$
  3. $2 \bar{a}+3 \bar{b}+4 \bar{c}$
  4. $\bar{a}+3 \bar{b}-4 \bar{c}$

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2017 (25 Apr Shift 1)

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