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Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow r. (\hat i + \hat j + \hat k) = 1 \: and \: \overrightarrow r. (2\hat i + 3\hat j - \hat k) +4 = 0$ and parallel to x - axis.

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  • Vector equation of a plane passing through the intersection of two planes is $\overrightarrow r.(\overrightarrow n_1+\lambda\overrightarrow n_2)=d_1+\lambda d_2$
  • If two planes are perpendicular,$a_1a_2+b_1b_2+c_1c_2=0$
Step 1:
Let the given planes be
$\overrightarrow r.(2\hat i+3\hat j-\hat k)+4=0$
$\overrightarrow r.(\hat i+\hat j+\hat k)-1=0$
Therefore equation of the plane passing through the line of intersection of these planes.
$\overrightarrow r.(2\hat i+3\hat j-\hat k)+\lambda[\overrightarrow r.(\hat i+\hat j+\hat k)-1]=0$
Expanding this we get
$\overrightarrow r.[(2+\lambda)\hat i+(3+\lambda)\hat j+(-1+\lambda)\hat k]+4-\lambda=0$------(1)
Step 2:
t is given the plane is parallel to the plane (1)
Its normal $\perp$ to $x$-axis.
Therefore the direction ratios of normal are $(2+\lambda,3+\lambda,-1+\lambda)$
The direction ratios of $x$-axis are $(1,0,0)$.
Normal of the plane and $x$-axis are $\perp$
Step 3:
Substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$(2+\lambda)\times 1+(3+\lambda)\times 0+(-1+\lambda)\times 0=0$
On simplifying we get
Therefore $\lambda=-2$
Step 4:
Put $\lambda=-2$ in equ(1)
$\overrightarrow r.[(2-2)\hat i+(3-2)\hat j+(-1-2)\hat k]+4+2=0$
Hence equation of required plane is $\overrightarrow r .(\hat j-3\hat k)+6=0$
answered Nov 14, 2013 by sreemathi.v

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