Question

If $\overrightarrow{a,}\overrightarrow{b}$are two non-collinear vectors, prove that the points with position vectors $\overrightarrow{a}+\overrightarrow{b,}\overrightarrow{a}-\overrightarrow{b}\mathrm{and}\overrightarrow{a}+\lambda \overrightarrow{b}$ are collinear for all real values of λ.

Answer 1

Given: $\overrightarrow{a},\overrightarrow{b}$ are non collinear vectors.
Let the position vectors of points A, B and C be $\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b},\overrightarrow{a}+\lambda \hat{b}$ respectively.
Then,
$\overrightarrow{AB}=$P.V. of B − P.V. of A.
$$\begin{gathered} =\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{b}. \\ =-2\overrightarrow{b}. \end{gathered}$$

$\overrightarrow{BC}=$P.V. of C − P.V. of B.
$$\begin{gathered} =\overrightarrow{a}+\lambda \overrightarrow{b}-\overrightarrow{a}+\overrightarrow{b}. \\ =\overrightarrow{b}\left(\lambda -1\right). \end{gathered}$$

$\overrightarrow{CA}=$P.V. of A − P.V. of C.
$$\begin{gathered} =\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{a}-\lambda \overrightarrow{b}. \\ =\overrightarrow{b}\left(1-\lambda \right). \end{gathered}$$
Now, the position vectors are collinear if and only if $\overrightarrow{AB}$ and $\overrightarrow{CA}$ is some multiple of $\overrightarrow{BC}$.
So,
$$\begin{gathered} \overrightarrow{AB}=\beta \stackrel{\rightharpoonup }{BC} \\ \Rightarrow -2b=\beta \stackrel{\rightharpoonup }{b}\left(\lambda -1\right) \\ \Rightarrow -2=\beta \left(\lambda -1\right) \\ \Rightarrow \beta =-\frac{2}{\lambda -1} \end{gathered}$$

and $\overrightarrow{BC}=-\overrightarrow{CA}.$
Hence, for real values of $\lambda$, the given position vectors are parallel.