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Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{1}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ And which is perpendicular to the plane $\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})+8=0$
Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{1}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ And which is perpendicular to the plane $\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})+8=0$
CBSE | Class 12 | Maths | Miscellaneous Exercise | NCERT | Three Dimensional Geometry
Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{1}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ And which is perpendicular to the plane $\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})+8=0$

Solution:

It is known that,
The eq. of any plane through the line of intersection of the planes $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{1}}=\mathrm{d}_{1}$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{2}}=\mathrm{d}_{2}$ is given by $\left(\vec{r} \cdot \overrightarrow{n_{1}}-\mathrm{d}_{1}\right)+\lambda\left(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}_{2}}-\mathrm{d}_{2}\right)^{2}=0$

Therefore, the eq. of any plane through the line of intersection of the given planes is

$\begin{array}{l}
{[\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4]+\lambda[\vec{r} \cdot(-2 \hat{i}-\hat{j}+\hat{k})-5]=0} \\
\vec{r} \cdot((1-2 \lambda) \hat{i}+(2-\lambda) \hat{j}+(3+\lambda) \hat{k})-4-5 \lambda=0 \\
\vec{r} \cdot((1-2 \lambda) \hat{i}+(2-\lambda) \hat{j}+(3+\lambda) \hat{k})=4+5 \lambda \\
\ldots(1)
\end{array}$

As this plane is perpendicular to the plane

$\begin{array}{l}
\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})+8=0 \\
\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})=-8 \\
-\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})=8 \\
\overrightarrow{\boldsymbol{r}} \cdot(-5 \hat{\boldsymbol{i}}-3 \hat{\boldsymbol{j}}+6 \hat{\boldsymbol{k}})=8
\end{array}$

Therefore, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).

The direction ratios of Normal of plane $(1)=\left(a_{1}, b_{1}, c_{1}\right) \equiv[(1-2 \lambda),(2-\lambda),(3+\lambda)]$

The direction ratios of Normal of plane (2) $=\left(\mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2}\right) \equiv(-5,-3,6)$

As the two lines are perpendicular,

$\begin{array}{l}
a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\
(1-2 \lambda) \times(-5)+(2-\lambda) \times(-3)+(3+\lambda) \times 6=0 \\
-5+10 \lambda-6+3 \lambda+18+6 \lambda=0 \\
19 \lambda+7=0 \\
\lambda=-7 / 19
\end{array}$

Substituting the value of $\lambda$ in eq.(1), we obtain

$\begin{array}{l}
\vec{r} \cdot((1-2 \lambda) \hat{i}+(2-\lambda) \hat{j}+(3+\lambda) \hat{k})=4+5 \lambda \\
\vec{r} \cdot\left(\left(1-2\left(\frac{-7}{19}\right)\right) \hat{i}+\left(2-\left(\frac{-7}{19}\right)\right) \hat{j}+\left(3+\left(\frac{-7}{19}\right)\right) \hat{k}\right)=4+5\left(\frac{-7}{19}\right) \\
\vec{r} \cdot\left(\frac{33}{19} \hat{i}+\frac{45}{19} \hat{j}+\frac{50}{19} \hat{k}\right)=\frac{41}{19} \\
\vec{r} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{k})=41
\end{array}$

As a result, the equation of the required plane is $\vec{r} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{k})=41$

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