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Find the equation of the plane which contains the line of intersection of the planes $\hat{r} . (\hat{i} + 2\hat{j} +3\hat{k}) - 4 = 0$ and $\hat{r} . (2\hat{i} + \hat{j} - \hat{k}) + 5 = 0$ and which is perpendicular to the plane $\hat{r} . (5\hat{i} + 3\hat{j} -6\hat{k}) +8 = 0$

$\begin{array}{1 1} (A) \overrightarrow r.(33\hat i+45\hat j+50\hat k)=0 \\(B) \overrightarrow r.(33\hat i+45\hat j+50\hat k)-41=0 \\ (C) \overrightarrow r.(33\hat i+45\hat j+50\hat k)+41=0 \\(D) \text{None of the above} \end{array} $

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  • If two planes are perpendicular to each other then $a_1a_2+b_1b_2+c_1c_2=0$
  • Equation of a plane passing through the intersection of two planes is $(\overrightarrow n_1+\lambda\overrightarrow n_2)=d_1+\lambda d_2$
Step 1:
Let the given planes be :
$\overrightarrow r.(\hat i+2\hat j+3\hat k)-4=0$-----(1)
$\overrightarrow r.(2\hat i+\hat j-\hat k)+5=0$-----(2)
A plane which contains the line of intersection of the planes (1) and (2) is
$\overrightarrow r.(\hat i+2\hat j+3\hat k)-4+\lambda[\overrightarrow r.(2\hat i+\hat j-\hat k)+5=0$
On simplifying we get
$\overrightarrow r.[(1+2\lambda)\hat i+(2+\lambda)\hat j+(3-\lambda)\hat k]-4+5\lambda=0$----(3)
Step 2:
Now the plane (3) is $\perp$ to the plane
$\overrightarrow r.(5\hat i+3\hat j-6\hat k)+8=0$
We know that $a_1a_2+b_1b_2+c_1c_2=0$
$\Rightarrow (1+2\lambda)\times 5+(2+\lambda)\times 3+(3-\lambda)\times (-6)=0$
On simplifying we get
Step 3:
Now substitute the value of $\lambda$ in equ(3) we get
$\overrightarrow r.[(1+\large\frac{14}{19})$$\hat i+(2+\large\frac{7}{19})$$\hat j+(3-\large\frac{7}{19})$$\hat k]$$-4+5\times \large\frac{7}{19}$$=0$
$\overrightarrow r.[\large\frac{33}{19}$$\hat i+\large\frac{45}{19}$$\hat j+\large\frac{50}{19}$$\hat k]+\large\frac{-76+35}{19}$$=0$
On simplifying we get,
$ \overrightarrow r.(33\hat i+45\hat j+50\hat k)-41=0$
This is the equation of the required plane.
answered Jun 5, 2013 by sreemathi.v

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