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Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7, \vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and through the point $(2,1,3)$
Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7, \vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and through the point $(2,1,3)$
CBSE | Class 12 | Exercise 11.3 | Maths | NCERT | Three Dimensional Geometry
Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7, \vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and through the point $(2,1,3)$

Solution:

Let’s consider the vector eq. of the plane passing through the intersection of the planes are
$\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=7 \text { and } \overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=9$
Here,
$\begin{array}{l}
\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})-7=0 \\
\overrightarrow{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-9=0 \\
\end{array}$
The eq. of any plane through the intersection of the planes given in eq. (1) and eq.(2) is given by,
$\begin{array}{l}
{[\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})-7]+\lambda[\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9]=0} \\
\vec{r}[(2 \hat{i}+2 \hat{j}-3 \hat{k})+(2 \lambda \hat{i}+5 \lambda \hat{j}+3 \lambda \hat{k})]-7-9 \lambda=0
\end{array}$
$\vec{r} \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(-3+3 \lambda) \hat{k}]-7-9 \lambda=0$
As the plane passes through points $(2,1,3)$
$\begin{array}{l}
(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot[(2+2 \lambda) \hat{\mathrm{i}}+(2+5 \lambda) \hat{\mathrm{j}}+(-3+3 \lambda) \hat{\mathrm{k}}]-7-9 \lambda=0 \\
4+4 \lambda+2+5 \lambda-9+9 \lambda-7-9 \lambda=0 \\
9 \lambda=10 \\
\lambda=10 / 9
\end{array}$
Now, on substituting $\lambda=10 / 9$ in eq.(1) we obtain,
$\vec{r} \cdot\left[\left(2+\frac{20}{9}\right) \hat{i}+\left(2+\frac{50}{9}\right) \hat{j}+\left(-3+\frac{30}{9}\right) \hat{k}\right]-7-9 \frac{10}{9}=0$
$\begin{array}{l}
\overrightarrow{\mathrm{r}} \cdot\left[\left(2+\frac{20}{9}\right) \hat{i}+\left(2+\frac{50}{9}\right) \hat{j}+\left(-3+\frac{30}{9}\right) \hat{k}\right]-17=0 \\
\vec{r} \cdot\left[\left(2+\frac{20}{9}\right) \hat{i}+\left(2+\frac{50}{9}\right) \hat{j}+\left(-3+\frac{30}{9}\right) \hat{k}\right]=17
\end{array}$
$\overrightarrow{\mathrm{r}}\left[\frac{38}{9} \hat{\mathrm{i}}+\frac{68}{9} \hat{\mathrm{j}}+\frac{3}{9} \hat{\mathrm{k}}\right]=17$
$\overrightarrow{\mathrm{r}}[38 \hat{\mathrm{i}}+68 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}]=153$
As a result, the required equation of the plane is $\overrightarrow{\mathrm{r}}[38 \hat{\mathrm{i}}+68 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}]=153$

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