Question

Match the elements of List I with their behaviour in List II:

Answer 1

(A) $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$

so $A\left(\overrightarrow{a}\right)=(1,2,3)$ and $\overrightarrow{b}=2\hat{i}+3\hat{j}+4\hat{k}$

and $L_2:\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-5}{5}=\mu$

so $C\left(\overrightarrow{c}\right)=(1,3,5)$ and $\overrightarrow{d}=3\hat{i}+4\hat{j}+5\hat{k}$

we find $\left[\begin{array}{ccc}\overrightarrow{c}-\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]=\left|\begin{array}{ccc}0& 1& 2\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=0$

so the lines are intersecting in a point

(B) $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$

and $L_2:\frac{x-3}{2}=\frac{y-5}{3}=\frac{z-7}{4}=\mu$

observe that $D.R^{\prime }$s are identical also for $\lambda =1$ and $\mu =0$ we get $(x,y,z)=(3,5,7)$ so lines are coincident (superimposition)

(C) $L_1:\frac{x-2}{5}=\frac{y+3}{4}=\frac{5-z}{-2}=\lambda$

and $L_2:\frac{x-7}{5}=\frac{y+1}{4}=\frac{z-2}{-2}=\mu$

observe that $D.R^{\prime }$s are identical

If the lines are coincident then $5\lambda +2=5\mu +7$ ...(i)

$4\lambda -3=4\mu -1$ ...(ii)

and $-2\lambda +5=-2\mu +2$ ....(iii)

Now from (i) and (ii) we get $\lambda -\mu =3$

and from (i) and (ii) we get $3(\lambda -\mu )=2$

so the lines do not intersect $\Rightarrow$lines are parallel and distinct

(D) $L_1:\frac{x-3}{2}=\frac{y+2}{4}=\frac{z-4}{5}=\lambda$

and $L_2:\frac{x-3}{3}=\frac{y-2}{2}=\frac{z-7}{5}=\mu$

so $\overrightarrow{a}=(3,-2,4);\overrightarrow{b}=(2,4,5);\overrightarrow{c}=(3,2,7);\overrightarrow{d}=(3,2,5)$

Now $\left[\begin{array}{ccc}\overrightarrow{c}-\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]=\left|\begin{array}{ccc}0& 4& 3\\ 2& 4& 5\\ 3& 2& 5\end{array}\right|=20-24=-4$

since$\left[\begin{array}{ccc}\overrightarrow{c}-\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]\ne 0$
skew lines