Question

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

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

Answer 1

Answer: (B)

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