Question

If the lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$, $\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}$ and $\frac{x-a}{3}=\frac{y-1}{2}=\frac{z-2}{b}$ are concurrent then:

• A. $a=-2,b=-6$
• B. $a=-\frac{1}{2},b=2$
• C. $a=-2,b=\frac{1}{2}$
• D. None

Answer 1

The correct option is B $a=-\frac{1}{2},b=2$

Given,

$L_1=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$

$L_2=\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}$

$L_3=\frac{x-a}{3}=\frac{y-1}{2}=\frac{z-2}{b}$

We know that if these lines intersect, there will be a common point of intersection.

Now,

If we put $L_1=k$

we get,

$L_1=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k$

$\therefore x=k,y=2k,z=3k$

If we put $L_2=h$

we get,

$L_2=\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}=h$

$\therefore x=3h+1,y=2-h,z=4h+3$

If we put $L_3=g$

we get,

$L_3=\frac{x-a}{3}=\frac{y-1}{2}=\frac{z-2}{b}=g$

$\therefore x=3g+a,y=2g+1,z=gb+2$

As we have one point of intersection, we can compare the coefficients of $L_1,L_2,L_3$

Comparing x coefficients of $L_{1}and{L}_2$

$\Rightarrow k=3h+1$

$\Rightarrow k-3h=\mathrm{1......}\left(1\right)$

Comparing y coefficients of $L_{1}and{L}_2$

$\Rightarrow 2k=2-h$

$\Rightarrow 2k+h=\mathrm{2......}\left(2\right)$

Multiplying eq(1) by 2 and substracting from eq(2), we get

$\Rightarrow 2k-6h-2k+h=2-2$

$\Rightarrow h=0$

Putting the value of h in eq(2), we get

$\Rightarrow 2k+0=2$

$\Rightarrow k=1$

Comparing y coefficients of $L_{2}and{L}_3$

$\Rightarrow 2-h=2g+1$

$\Rightarrow 2=2g+1$

$\Rightarrow g=\frac{1}{2}$

Comparing x coefficients of $L_{2}and{L}_3$

$\Rightarrow 3h+1=3g+a$

$\Rightarrow 1=3\left(\frac{1}{2}\right)+a$

$\Rightarrow 1-\frac{3}{2}=a$

$\Rightarrow -\frac{1}{2}=a$

Comparing z coefficients of $L_{2}and{L}_3$

$\Rightarrow 4h+3=gb+2$

$\Rightarrow 3=\frac{b}{2}+2$

$\Rightarrow 3-2=\frac{b}{2}$

$\Rightarrow b=2$