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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane passing through (a, b, c) and parallel to the plane $\hat{r}. (\hat{i}+\hat{j}+\hat{k}) = 2$.

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  • The plane parallel to another given plane $\overrightarrow r.\overrightarrow n_1=\overrightarrow d_1$ is $\overrightarrow r.\overrightarrow n_1=\overrightarrow d_2$
Step 1:
Let the plane parallel to the plane $\overrightarrow r_1.(\hat i+\hat j+\hat k)=2$ be of the form $\overrightarrow r.(\hat i+\hat j+\hat k)=\lambda$------(1)
It is given the plane passes through the point $(a,b,c)$
Therefore let the position vector $\overrightarrow r$ of this point be $\overrightarrow r=a\hat i+b\hat j+c\hat k$
Step 2:
Substituting for $\overrightarrow r$ in equ(1) we get
$(a\hat i+b\hat j+c\hat k).(\hat i+\hat j+\hat k)=\lambda$
Applying the scalar product rule and multiplying we get
Step 3:
Substituting for $\lambda$ we get,
$\overrightarrow r.(\hat i+\hat j+\hat k)=a+b+c$
This is the vector equation of the required plane.
But we know $\overrightarrow r=x\hat i+y\hat j+z\hat k$
Therefore $(x\hat i+y\hat j+z\hat k).(\hat i+\hat j+\hat k)=a+b+c$
Step 4:
We know $\hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1$
On multiplying we get
answered Jun 4, 2013 by sreemathi.v

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