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Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVectors2 Marks
MCQ
If the three points with position vectors $\bar{a}-2 \bar{b}+3 \bar{c}, 2 \bar{a}+\lambda \bar{b}-4 \bar{c}, -7 \bar{b}+10 \bar{c} $ are collinear, then $\lambda=$
  • A
    1
  • B
    2
  • ✓
    3
  • D
    none of these
✓

Answer

Correct option: C.
3
(C) Let $\overline{ p }=\overline{ a }-2 \overline{b}+3 \overline{ c }, \overline{ q }=2 \overline{ a }+\lambda \overline{ b }-4 \overline{ c } $ and $\overline{ r }=-7 \overline{b}+10 \overline{ c }$
Since points are collinear
$\therefore \quad \overline{ PQ }= k \overline{ PR }$
$\Rightarrow 2 \overline{ a }+\lambda \overline{ b }-4 \overline{ c }-(\overline{ a }-2 \overline{b}+3 \overline{ c })$
$= k [-7 \overline{b}+10 \overline{ c }-(\overline{ a }-2 \overline{b}+3 \overline{ c })]$
$\Rightarrow \overline{ a }+(\lambda+2) \overline{ b }-7 \overline{ c }=- k \overline{ a }-5 k \overline{ b }+7 k \overline{ c }$
$\Rightarrow k =-1, \lambda=-5 k -2$
Hence, $\lambda=5-2=3$

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