Question

The vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-collinear. Value of $x$ ,for the vectors $c=(x-2)a+b$ and $d=(2x+1)a-b$ are collinear

• A. $x=-\frac{1}{6}.$
• B. $x=+\frac{1}{6}.$
• C. $x=-\frac{1}{3}.$
• D. $x=+\frac{1}{3}.$

Answer 1

Answer: (D)

$x=+\frac{1}{3}.$

Both the vectors c and d are non-zero as the coefficients of b in both are non-zero. Two vectors c and d are collinear if one of them is a linear multiple of the other.
$\therefore d=\lambda c$
or $(2x+1)a-b=\lambda \{(x-2)a+b\}$...(1) or or $\{(2x+1)-\lambda (x-2)\}a-(1+\lambda )b=0$
where a and b are non-collinear and hence we must have coefficients of $a$ and $b$ zero.
$\therefore 2x+1-\lambda (x-2)=0$ ...(2)
$1+\lambda =0$ ...(3)
From (2), $\lambda =-1$ and putting in (1) we get $x=\frac{1}{3}.$
Note : We could also say compare the coefficients of a and b (non-collinear) in (1) and we get the results (2) and (3).