Question

If $a , b , c$ are non coplanar vectors, then the point of intersection of the line passing through the points $2 a +3 b - c , 3 a +4 b -2 c$ with the line joining the points $a -2 b +3 c$, $a -6 b +6 c$ is

• A. $a + b + c$
• B. $a +2 b$
• C. $a + c$
• D. $\frac{a+2 b+c}{2}$

Answer 1

Answer: (B)

Let, $O A =2 a +3 b - c , O B =3 a +4 b -2 c$
$O C = a -2 b +3 c$, and $O D = a -6 b +6 c$
The vector equation of line joining the points $A$ and $B$ is
$r = O A +t( O B - O A ), t \in R$
$=(2 a +3 b - c )+t(3 a +4 b -2 c )-(2 a +3 b - c )]$
$r =2 a +3 b - c +t( a + b - c )\,\,\,...(i)$
$=(2+t) a +(3+t) b +(-1-t) c\,\,\,...(ii)$
Vector equation of the line joining the points $C$ and $D$ is
$r = O C +s( O D - O C )$
$= a -2 b +3 c +s( a -6 b +6 c )-( a -2 b +3 c )$
$r = a +(-2-4 s) b +(3+3 s) c \,\,\,\ldots(iii)$
Comparing coefficient of a in Eq. (ii) and (iii)
' we get $2+t=1 $
$\Rightarrow t=-1$
Put in Eq. (i)
$r =2 a +3 b - c +(-1)[ a + b - c ]$
$=2 a +3 b - c - a - b + c$
$r = a +2 b$
So, the point of in intersection is $a +2 b$