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Introduction to Three Dimensional Geometry — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsIntroduction to Three Dimensional Geometry1 Mark
MCQ
A point $C$ with position vector $\frac{\text{3a}+4\text{b}-5\text{c}}{3} ($where $a, b$ and $c$ are non co$-$planar vectors$)$ divides the line joining $A$ and $B$ in the ratio $2 : 1.$ If the position vector of $A$ is $a - 2b + 3c,$ then the position vector of $B$ is:
  • A
    $2a + 3b - 4c$
  • B
    $2a - 3b + 4c$
  • C
    $2a + 3b + 4c$
  • ✓
    $a + 3b - 4c$
✓

Answer

Correct option: D.
$a + 3b - 4c$
$\frac{\text{3a}+4\text{b}-5\text{c}}{3}$
$\overrightarrow{\text{c}}=\frac{2\overrightarrow{\text{b}}+\overrightarrow{\text{a}}}{3}$
$\overrightarrow{\text{b}}=\frac{3\overrightarrow{\text{c}}-\overrightarrow{\text{a}}}{2}$
$=\frac{\big(3\overrightarrow{a}+4\overrightarrow{b}-5\overrightarrow{c}\big)-\big(\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}\big)}{2}$
$=\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$

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