CBSE 2023 · Region 4 · Set 1 · Q35 · 5 marks
Show that the following lines do not intersect each other : \[\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} ; \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2} \]Find the angle between the lines \[2 x=3 y=-z \text { and } 6 x=-y=-4 z \]
Show that the following lines do not intersect each other : \[\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} ; \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2} \]
Find the angle between the lines \[2 x=3 y=-z \text { and } 6 x=-y=-4 z \]
Marking-scheme solution
Shortest Distance $\displaystyle = \dfrac{\begin{vmatrix} -3 & 2 & -2 \\ 3 & 2 & 5 \\ 4 & 3 & -2 \end{vmatrix}}{\sqrt{(2\times-2-5\times3)^2+(3\times-2-4\times5)^2+(3\times3-2\times4)^2}}$
$\displaystyle = \dfrac{-3(-19)-2(-26)-2(1)}{\sqrt{361+676+1}}$
$\displaystyle = \dfrac{57+52-2}{\sqrt{1038}}$
$\displaystyle = \dfrac{107}{\sqrt{1038}} \neq 0$
So, the line will not intersect each other.
The given lines are
$\displaystyle \dfrac{x-0}{1/2} = \dfrac{y-0}{1/3} = \dfrac{z-0}{-1}$ and $\displaystyle \dfrac{x-0}{1/6} = \dfrac{y-0}{-1} = \dfrac{z-0}{-1/4}$
Or $\displaystyle \dfrac{x}{3} = \dfrac{y}{2} = \dfrac{z}{-6}$ and $\displaystyle \dfrac{x}{2} = \dfrac{y}{-12} = \dfrac{z}{-3}$
Let $\displaystyle \theta$ be the angle between the two lines, then.
$\displaystyle \cos\theta = \left| \dfrac{(3\times2) + (2\times-12) + (-6)(-3)}{\sqrt{9+4+36}\ \sqrt{4+144+9}} \right|$
$\displaystyle = \left| \dfrac{6-24+18}{7\times\sqrt{157}} \right|$
$\displaystyle = 0$
$\displaystyle \Rightarrow \theta = 90°$
Three Dimensional GeometryShortest Distance between Two LinesApplylong_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.