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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the vector equation of the plane passing through the intersection of the planes $ \overrightarrow r .(2\hat i + 2 \hat j - 3\hat k) = 7, \overrightarrow r .(2\hat i + 5\hat j + 3\hat k ) = 9$ and through the point $(2, 1, 3).$

$\begin{array}{1 1} \overrightarrow r .(4\hat i + 7\hat j -15\hat k)=30 \\ \overrightarrow r .(4\hat i + 7\hat j -15\hat k)=153 \\ \overrightarrow r .(38\hat i + 68\hat j + 3\hat k)=153 \\ \overrightarrow r .(38\hat i + 68\hat j + 3\hat k)=-153\end{array} $

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  • Vector equation of a plane passing through the intersection of two planes is $\overrightarrow r.(\overrightarrow {n_1}+\lambda\overrightarrow {n_2})=\overrightarrow d_1+\lambda \overrightarrow d_2$
Step 1:
Given vector equations of the plane are
$\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7$
$\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9$
These equations can be written as
$\overrightarrow r.(2\hat i+2\hat j-3\hat k-7)=0$
$\overrightarrow r.(2\hat i+5\hat j+3\hat k-9)=0$
Equation of any plane passing through the intersection of the given planes is given by
$[\overrightarrow r.(2\hat i+2\hat j-3\hat k-7)]+\lambda[\overrightarrow r.(2\hat i+5\hat j+3\hat k-9)]=0$
$\Rightarrow [\overrightarrow r.(2\hat i+2\hat j-3\hat k)]+\lambda[\overrightarrow r.(2\hat i+5\hat j+3\hat k)]=7+9\lambda$----(1)
This passes through the point $(2,1,3)$
Therefore its position vector is $\overrightarrow r=2\hat i+\hat j+3\hat k$
Step 2:
Hence substituting for $\overrightarrow r$ in equ(1) we get,
$[(2\hat i+\hat j+3\hat k).(2\hat i+2\hat j-3\hat k)]+\lambda[(2\hat i+\hat j+3\hat k).(2\hat i+5\hat j+3\hat k)]=7+9\lambda$
$\Rightarrow 2(2+2\lambda)+1(2+5\lambda)+3(3\lambda-3)=7+9\lambda$
(i.e)$4+4\lambda+2+5\lambda-9+9\lambda=7+9\lambda$
On simplifying we get
$9\lambda-3=7$
$9\lambda=10$
$\lambda=\large\frac{10}{9}$
Step 3:
Substituting the value of $\lambda$ in equ(1) we get
$[\overrightarrow r.(2\hat i+2\hat j-3\hat k)]+\large\frac{10}{9}$$[\overrightarrow r.(2\hat i+5\hat j+3\hat k)]=7+9.\large\frac{10}{9}$
On simplifying we get
$\Rightarrow \overrightarrow r\big(\large\frac{38}{9}$$\hat i+\large\frac{68}{9}$$\hat j+\large\frac{3}{9}$$\hat k\big)=17$
Therefore $\overrightarrow r.(38\hat i+68\hat j+3\hat k)=153$
This is the vector equation in the required plane.
answered May 31, 2013 by sreemathi.v
 

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