Question

If $$\overrightarrow{a,}\overrightarrow{b}$$ are two non-collinear vectors prove that the points with position vectors $$\overrightarrow{a}+\overrightarrow{b,}\overrightarrow{a}-\overrightarrow{b}\text{ and }\overrightarrow{a}+\lambda \overrightarrow{b}$$ 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 $$\overrightarrow{a}+\overrightarrow{b,}\overrightarrow{a}-\overrightarrow{b},\overrightarrow{a}+\lambda \overrightarrow{b}$$  respectively.
Then, $$\overrightarrow{AB}=$$ P.V. of B − P.V. of A. 
$$=\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{a}-\overrightarrow{b}.$$
$$=-2\overrightarrow{b}.$$
$$\overrightarrow{BC}=$$ P.V. of C − P.V. of B.
$$=\overrightarrow{a}+\lambda \overrightarrow{b}-\overrightarrow{a}+\overrightarrow{b}.$$
$$=\overrightarrow{b}(\lambda -1).$$
$$\overrightarrow{CA}=$$ P.V. of A − P.V. of C.
$$=\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{a}-\lambda \overrightarrow{b}.$$
$$=\overrightarrow{b}(1-\lambda ).$$
Now, the position vectors are collinear if and only if $$\overrightarrow{AB}$$ and $$\overrightarrow{CA}$$  is some multiple of $$\overrightarrow{BC}$$
So,
$$\overrightarrow{AB}=\beta \overrightarrow{BC}$$
$$\Rightarrow -2b=\beta \overrightarrow{b}(\lambda -1)$$
$$\Rightarrow -2=\beta (\lambda -1)$$
$$\Rightarrow \beta =-\frac{2}{\lambda -1}$$ and $$\overrightarrow{BC}=-\overrightarrow{CA}.$$
Hence, for real values of $$\lambda$$, the given position vectors are parallel.