Question

Three points whose position vectors are $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ will be collinear if

• A. $\lambda \overrightarrow{a}+\mu \overrightarrow{b}=(\lambda +\mu )\overrightarrow{c}$
• B. $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}$
• C. $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=0$
• D. None of these

Answer 1

Answer: (A), (B)

$\lambda \overrightarrow{a}+\mu \overrightarrow{b}=(\lambda +\mu )\overrightarrow{c}$
$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}$

If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are collinear vectors, therefore they lie on the same line. Hence they are parallel. Hence there cross products will be 0.
Therefore
$(\overrightarrow{a}\times \overrightarrow{b})=(\overrightarrow{c}\times \overrightarrow{b})=(\overrightarrow{a}\times \overrightarrow{c})=0$
And application of the section formula gives us
$\overrightarrow{c}=\frac{\lambda \overrightarrow{a}+\mu \overrightarrow{b}}{\lambda +\mu }$
Or
$\overrightarrow{c}(\lambda +\mu )=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$