Check whether the lines x−12=y−23=z−34 and x−45=y−12=z are skew or not.

Given: x−12=y−23=z−34……(1) and x−45=y−12=z−01........(2)So, x1=1,y1=2,z1=3 and x2=4,y2=1,z2=0 also l1=2,m1=3,n1=4 and l2=5,m2=2,n2=1Now for to check the lines are skew then it should not be coplanar,So, for coplanarity,x2−x1y2−y1z2−z1l1m1n1l2m2n2=0Now, x2−x1y2−y1z2−z1l1m1n1l2m2n2=4−11−20−3234521⇒3−1−3234521=3(3−8)−(−1)(2−20)−3(4−15)=3(−5)−18−3(−11)=−15−18+33=−33+33=0Therefore, we can say that lines are coplanar. Hence, given lines are not skew.

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Check whether the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 4}{5} = \frac{y - 1}{2} = z$ are skew or not.

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Detailed Solution

Given: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \dots \dots (1)$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1} . . . . . . . . (2)$
So, $x_{1} = 1, y_{1} = 2, z_{1} = 3$ and $x_{2} = 4, y_{2} = 1, z_{2} = 0$ also $l_{1} = 2, m_{1} = 3, n_{1} = 4$ and $l_{2} = 5, m_{2} = 2, n_{2} = 1$
Now for to check the lines are skew then it should not be coplanar,
So, for coplanarity,
$|\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}| = 0$
Now,
$\begin{array}{l} |\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}| = |\begin{array}{ccc} 4 - 1 & 1 - 2 & 0 - 3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{array}| \\ \Rightarrow |\begin{array}{ccc} 3 & - 1 & - 3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{array}| \\ = 3 (3 - 8) - (- 1) (2 - 20) - 3 (4 - 15) \\ = 3 (- 5) - 18 - 3 (- 11) \\ = - 15 - 18 + 33 = - 33 + 33 = 0 \end{array}$
Therefore, we can say that lines are coplanar. Hence, given lines are not skew.