Question

Let $a,b$ and $c$ be three non-coplanar vectors. The vector equation of a line which passes through the point of intersection of two lines, one joining the points $a+2b-5c$, $-a-2b-3c$ and the other joining the points $-4c,6a-4b+4c$ is

• A. $r=2a-4b+3c+\mu (a-6b+4c)$
• B. $r=3a+6b-c+\mu (a+2b+c)$
• C. $r=2a+3b-c+\mu (a+b-c)$
• D. $r=-2b+3c+\mu (a-4b+3c)$

Answer 1

Answer: (B)

Equation of line joining the points,
$a+2b-5c,-a-2b-3c$ is
$r=(a+2b-5c)+{\lambda }_1,(2a+4b-2c)$ .....(i)
Similarly, equation of the line joining the points
$-4c,6a-4b+4c$ is
$r=-4c+{\lambda }_2(6a-4b+8c)$ ......(ii)
Now, for point of intersection of lines (i) and (ii)
$\left(2{\lambda }_1+1\right)a+\left(4{\lambda }_1+2\right)b+(-2\lambda ,-5)c$
$=\left(6{\lambda }_2\right)a+\left(-4{\lambda }_2\right)b+\left(8{\lambda }_2+4\right)c$
On comparing
$2{\lambda }_1+1=6{\lambda }_2$....(iii)
$4{\lambda }_1+2=-4{\lambda }_2$......(iv)
and $-2{\lambda }_1-5=8{\lambda }_2+\mathrm{4........}(v)$
From Eqs. (iii), (iv) and (v)
${\lambda }_1=-\frac{1}{2}\text{ and }{\lambda }_2=0$
So, intersection points is $-4c$ and for $\mu =-3$ the
line $r=(3a+6b-c)+\mu (a+2b+c)$ passes through
point $-4c$ only.