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 $a +2 \,b -5\,c$, $- a -2 b -3 c$ and the other joining the points $-4\,c, \, 6 \,a -4\, b +4\, c$ is

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

Answer 1

Answer: (B)

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