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Home » Find the equations of the line passing through the point (-1, 3, -2) and perpendicular to each of the lines and
Find the equations of the line passing through the point (-1, 3, -2) and perpendicular to each of the lines and
CBSE | Class 12 | Exercise 27A | Maths | RS Aggarwal | Straight Line in Space
Find the equations of the line passing through the point (-1, 3, -2) and perpendicular to each of the lines and

Find the equations of the line passing through the point (-1,3,-2) and perpendicular to each of the lines \frac{\mathrm{X}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{Z}}{3} and \frac{\mathrm{x}+2}{-3}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+1}{5}
Answer
Given: line passes through (-1,3,-2) and is perpendicular to each of the lines \frac{x}{1}=\frac{y}{2}=\frac{\pi}{3} and \frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}
To find: equation of line in Vector and Cartesian form
Formula Used: Equation of a line is
Vector form: \overrightarrow{\mathrm{I}}=\overrightarrow{\mathrm{d}}+\lambda \overrightarrow{\mathrm{b}}
Cartesian form: \frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{z}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{~h}_{\mathrm{g}}}=\lambda.
where \vec{a}=\mathrm{z}_{1} \hat{\mathrm{l}}+\mathrm{y}_{1} \hat{l}+\mathrm{z}_{1} \hat{\mathrm{k}} is a point on the line and \overrightarrow{\mathrm{b}}=\mathrm{h}_{1} \hat{\mathrm{i}}+\mathrm{h}_{2} \hat{\mathrm{l}}+\mathrm{b}_{3} \hat{\mathrm{k}} is a vector parallel to the line.
If 2 lines of direction ratios a_{1}: a_{2}: a_{3} and b_{1}: b_{2}: b_{3} are perpendicular, then a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}=0
Explanation:
Here, \vec{A}=-\hat{\mathrm{I}}+\hat{\mathrm{a}} \hat{\mathrm{i}}-2 \hat{\mathrm{k}}
Let the direction ratios of the line be \mathrm{b}_{1}: \mathrm{b}_{2}: \mathrm{b}_{3}
Direction ratios of the other two lines are 1: 2: 3 and -3: 2: 5
Since the other two line are perpendicular to the given line, we have

Solving,
\frac{\|_{3}}{\left|\begin{array}{ll}2 & 3 \\ 2 & 5\end{array}\right|}=\frac{-b_{2}}{\left|\begin{array}{cc}1 & 3 \\ -3 & 5\end{array}\right|}=\frac{5_{3}}{\left|\begin{array}{cc}1 & 2 \\ -3 & 2\end{array}\right|}
\Rightarrow \frac{\mathrm{b}_{1}}{\mathrm{q}}=\frac{\mathrm{b}_{2}}{-\mathrm{J} . \mathrm{k}}=\frac{\mathrm{b}_{\mathrm{j}}}{8 \mathrm{z}}
\Rightarrow \frac{\mathrm{b}_{1}}{\mathrm{z}_{1}}=\frac{\mathrm{b}_{2}}{-7}=\frac{\mathrm{b}_{\mathrm{z}}}{\mathrm{i}_{\mathrm{k}}}

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