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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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If the lines $\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(3\hat i+2k\hat j-2\hat k)$ and $\overrightarrow r=(\hat i+2\hat j+3\hat k)+\mu(k\hat i+\hat j+5\hat k)$ are $\perp$ to one another then find the equation of the plane containing them.

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Toolbox:
  • If two lines $\overrightarrow r=\overrightarrow a_1+\lambda\overrightarrow b_1$ and $\overrightarrow r=\overrightarrow a_2+\mu \overrightarrow b_2$ are $\perp$ then $\overrightarrow b_1.\overrightarrow b_2=0$
  • The equation of the plane containing the lines $\large\frac{x-x_1}{l_1}$$=\large\frac{y-y_1}{m_1}$$=\large\frac{z-z_1}{n_1}$$=\lambda$ and $\large\frac{x-x_2}{l_2}$$=\large\frac{y-y_2}{m_2}$$=\large\frac{z-z_2}{n_2}$$=\mu$is given by $\left|\begin {array}{ccc} x-x_1 & y-y_1 &z-z_1\\ l_1 & m_1 & n_1\\ l_2 &m_2 &n_2\end {array} \right|=0$
Step 1
The equations of the given lines in cartesian form are
$\large\frac{x-1}{3}$$=\large\frac{y-2}{2k}$$=\large\frac{z-3}{-2}$$=\lambda$...........(i) and
$\large\frac{x-1}{k}$$=\large\frac{y-2}{1}$$=\large\frac{z-3}{5}$$=\mu$......(ii)
$\Rightarrow\:$ The direction ratios $(d.r)$ of the lines are $\overrightarrow b_1=(3,2k,-2)$ and $\overrightarrow b_2=(k,1,5)$
Given that the lines (i) and (ii) are $\perp$
$\therefore\:\overrightarrow b_1.\overrightarrow b_2=0$
$\Rightarrow$ $(3,2k,-2).(k,1,5)=0$
$\Rightarrow\:3k+2k-10=0\:\:or\:\:k=2$
$\therefore$ By substituting the value of $k$ in (i) and (ii) the given equations of the lines become
$\large\frac{x-1}{3}$$=\large\frac{y-2}{4}$$=\large\frac{z-3}{-2}$$=\lambda$...........(i) and
$\large\frac{x-1}{2}$$=\large\frac{y-2}{1}$$=\large\frac{z-3}{5}$$=\mu$......(ii)
Step 2
The equation of the plane containing the lines $\large\frac{x-x_1}{l_1}$$=\large\frac{y-y_1}{m_1}$$=\large\frac{z-z_1}{n_1}$$=\lambda$ and
$\large\frac{x-x_2}{l_2}$$=\large\frac{y-y_2}{m_2}$$=\large\frac{z-z_2}{n_2}$$=\mu$
is given by $\left|\begin {array}{ccc} x-x_1 & y-y_1 &z-z_1\\ l_1 & m_1 & n_1\\ l_2 &m_2 &n_2\end {array} \right|=0$
$\Rightarrow$ The equation of the required plane containing (i) and (ii) is given by
is given by $\left|\begin {array}{ccc} x-1 & y-2 &z-3\\ 3 & 4 & -2\\ 2 &1 & 5\end {array} \right|=0$
$\Rightarrow\:$ The equation is $22x-19y-5z+31=0$
answered Feb 4, 2014 by rvidyagovindarajan_1
 

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