
Find the equation of lines passing through the point \[\left( 3,1,2 \right)\]and perpendicular to the lines \[\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}\]and \[\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}\]
Answer
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Hint: we know that when two lines are perpendicular to each other then the condition for their direction ratios is \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\]. Now wee will get two equations and by cross multiplication we will get the direction ratios of the required line then we know the equation of line passing through a point and having direction ratios a, b, c.
Complete step-by-step answer:
Let a, b, c be the direction ratios of required ratios of the required line which is perpendicular to the lines \[\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}\]and \[\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}\]
We know that when two lines are perpendicular to each other then the condition for direction cosines is \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\]
\[a+2b+3c=0\]. . . . . . . . . . . . . . . . . . . (1)
\[-3a+2b+5c=0\].. . . . . . . . . . . . . . . . (2)
Do the cross multiplication then we will get
\[=\dfrac{a}{\left( \begin{matrix}
2 & 3 \\
2 & 5 \\
\end{matrix} \right)}=\dfrac{-b}{\left( \begin{matrix}
1 & 3 \\
-3 & 2 \\
\end{matrix} \right)}=\dfrac{c}{\left( \begin{matrix}
1 & 2 \\
-3 & 2 \\
\end{matrix} \right)}\]
\[=\dfrac{a}{4}=\dfrac{b}{-14}=\dfrac{c}{8}\]. . . . . . . . . . . . . . . . . . (3)
The direction ratios of required line is 4, -14, 8 or 2, -7, 4
The equation of a line passing through the point \[\left( {{x}_{1,}}{{y}_{1,}}{{z}_{1}} \right)\]and having direction ratios a, b, c is \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]
The equation of line passing through A \[\left( 3,1,2 \right)\]and having direction ratios 2, -7, 4 is
\[\dfrac{x-3}{2}=\dfrac{y-1}{-7}=\dfrac{z-2}{4}\]
Note: The equation of line passing through a point A \[\left( {{x}_{1,}}{{y}_{1,}}{{z}_{1}} \right)\]and having direction ratios a, b, c in cartesian form is \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]. If two lines are parallel to each other then the direction ratios of those two lines are equal.
Complete step-by-step answer:
Let a, b, c be the direction ratios of required ratios of the required line which is perpendicular to the lines \[\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}\]and \[\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}\]
We know that when two lines are perpendicular to each other then the condition for direction cosines is \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\]
\[a+2b+3c=0\]. . . . . . . . . . . . . . . . . . . (1)
\[-3a+2b+5c=0\].. . . . . . . . . . . . . . . . (2)
Do the cross multiplication then we will get
\[=\dfrac{a}{\left( \begin{matrix}
2 & 3 \\
2 & 5 \\
\end{matrix} \right)}=\dfrac{-b}{\left( \begin{matrix}
1 & 3 \\
-3 & 2 \\
\end{matrix} \right)}=\dfrac{c}{\left( \begin{matrix}
1 & 2 \\
-3 & 2 \\
\end{matrix} \right)}\]
\[=\dfrac{a}{4}=\dfrac{b}{-14}=\dfrac{c}{8}\]. . . . . . . . . . . . . . . . . . (3)
The direction ratios of required line is 4, -14, 8 or 2, -7, 4
The equation of a line passing through the point \[\left( {{x}_{1,}}{{y}_{1,}}{{z}_{1}} \right)\]and having direction ratios a, b, c is \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]
The equation of line passing through A \[\left( 3,1,2 \right)\]and having direction ratios 2, -7, 4 is
\[\dfrac{x-3}{2}=\dfrac{y-1}{-7}=\dfrac{z-2}{4}\]
Note: The equation of line passing through a point A \[\left( {{x}_{1,}}{{y}_{1,}}{{z}_{1}} \right)\]and having direction ratios a, b, c in cartesian form is \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]. If two lines are parallel to each other then the direction ratios of those two lines are equal.
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