Question

If the three points with position vectors $\overline{a}-2\overline{b}+3\overline{c},2\overline{a}+\lambda \overline{b}-4\overline{c},-7\overline{b}+10\overline{c}$ are collinear, then $\lambda =$

• A. $1$
• B. 2
• C. $3$
• D. none of these

Answer 1

Answer: (C)

$3$

The given vectors are collinear, so $l(\overline{a}-2\overline{b}+3\overline{c})+k(2\overline{a}+\lambda \overline{b}-4\overline{c})=(l+k)(-7\overline{b}+10\overline{c})$
Comparing the coefficients of $\overline{a}\to l+2k=0$
$\overline{b}\to -2l+\lambda k=-7l-7k$
$\overline{c}\to 3l-4k=10l+10k$
$\Rightarrow l=-2k$ and so $\lambda =3$