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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the vector equation of the line passing through (1, 2, -4) and perpendicular to the two lines $\frac{\large x-8}{\large 3} = \frac{\large y+19}{\large -16} = \frac{\large z-10}{\large 7}$ and $\frac{\large x-15}{\large 3} = \frac{\large y-29}{\large 8} = \frac{\large z-5}{\large -5}$

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1 Answer

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Toolbox:
  • Vector equation of a line passing through a given point and parallel to a given vector is $\overrightarrow r=\overrightarrow a+\lambda\overrightarrow b$
  • Where $\lambda\in R$
Step 1:
Let the vector equation of the line passing through the point $(1,2,-4)$ be
$\overrightarrow r=\hat i+2\hat j-4\hat k+\lambda(b_1\hat i+b_2\hat j+b_3\hat k)$------(1)
Where $b_1\hat i+b_2\hat j+b_3\hat k$ is the vector parallel to the given line.
Step 2:
Let us consider the lines $\large\frac{x-8}{3}=\large\frac{y+19}{-16}=\large\frac{z-10}{7}$-----(2)
The direction cosines of the lines are $(3,-16,7)$
The vector parallel to that line is $3\hat i-16\hat j+7\hat k$
But it is given the line (1) and (2) are perpendicular to each other.
Therefore $(b_1\hat i+b_2\hat j+b_3\hat k).(3\hat i-16\hat j+7\hat k)=0$
$\Rightarrow 3b_1-16b_2+7b_3=0$-----(3)
Step 3:
Consider the line:
$\large\frac{x-11}{3}=\large\frac{y-29}{8}=\large\frac{x-5}{-5}$----(4)
The vector which is parallel to the line is $5\hat i+3\hat j-5\hat k$
The lines (1) and (4) are perpendicular to each other.
Therefore $(b_1\hat i+b_2\hat j+b_3\hat k).(3\hat i+8\hat j-5\hat k)=0$
$\Rightarrow 3b_1+8b_2-5b_3=0$-----(5)
Step 4:
On solving equ(3) and equ(5)
$\large\frac{b_1}{\begin{vmatrix}-16 & 7\\8 & -5\end{vmatrix}}=\large\frac{b_2}{\begin{vmatrix}7 & 3\\-5 & 3\end{vmatrix}}=\large\frac{b_1}{\begin{vmatrix}3 & -16\\3& 8\end{vmatrix}}$
$\large\frac{b_1}{80-56}=\large\frac{b_2}{21+15}=\large\frac{b_3}{24+48}$
$\Rightarrow \large\frac{b_1}{24}=\large\frac{b_2}{36}=\large\frac{b_3}{72}$
Step 5:
Dividing throughout by 12
$\Rightarrow \large\frac{b_1}{2}=\large\frac{b_2}{3}=\large\frac{b_3}{6}$
Therefore direction of the vector which is parallel to line (1) is $2\hat i+3\hat j+6\hat k$
Therefore Equation of the line (1) is
$\overrightarrow r=\hat i+2\hat j-4\hat k+\lambda(2\hat i+3\hat j+6\hat k)$
answered Jun 6, 2013 by sreemathi.v
 

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