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The vector equation of any plane passing through the line of intersection of the planes $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{m}}_1=\mathrm{q}_1$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{m}}_2=\mathrm{q}_2$ is given by $\overrightarrow{\mathrm{r}}\left(\overrightarrow{\mathrm{m}}_1+\lambda \overrightarrow{\mathrm{m}}_2\right)=\mathrm{q}_1+\lambda \mathrm{q}_2$ for $\lambda \in \mathbb{R}$. The vector equation of a plane passing through the point $2 \hat{i}-3 \hat{j}+\hat{k}$ and the line of intersection of the planes r. $(\hat{i}-2 \hat{j}+3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}+\hat{j}-2 \hat{k})=7$
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Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=12$
Given, planes are
$$
\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=5
$$ ...(i)
and $\vec{r} \cdot(3 \hat{i}+\hat{j}-2 \hat{k})=7$ ...(ii)
Hence equation of plane passing through the line of intersection of the plane, (i) and plane (ii) is given by
$$
\begin{aligned}
& \vec{r}_1 \cdot\left(\vec{m}_1+\lambda \vec{m}_2\right)=q_1+\lambda q_2, \lambda \in R \\
\Rightarrow & \vec{r} \cdot[(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+\hat{j}-2 \hat{k})]=5+\lambda(7) \\
\Rightarrow & \vec{r} \cdot[\hat{i}(1+3 \lambda)+\hat{j}(-2+\lambda)+\hat{k}(3-2 \lambda)]=5+7 \lambda
\end{aligned}
$$ ...(iii)
$$
\begin{aligned}
& \text { but given that } \vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k}) \\
& \Rightarrow 2(1+3 \lambda)+(-3)(-2+\lambda)+(1)(3-2 \lambda)=5+7 \lambda \\
& \Rightarrow 11+\lambda=5+7 \lambda \Rightarrow 6 \lambda=6 \Rightarrow \lambda=1 .
\end{aligned}
$$
putting $\lambda=1$ in equation (iii), we get
$$
\begin{aligned}
& \Rightarrow \quad \vec{r} \cdot[\hat{i}(1+3)+\hat{j}(-2+1)+\hat{k}(3-2)]=5+7 \\
& \Rightarrow \quad \vec{r} \cdot[4 \hat{i}-\hat{j}+\hat{k}]=12
\end{aligned}
$$
$$
\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=5
$$ ...(i)
and $\vec{r} \cdot(3 \hat{i}+\hat{j}-2 \hat{k})=7$ ...(ii)
Hence equation of plane passing through the line of intersection of the plane, (i) and plane (ii) is given by
$$
\begin{aligned}
& \vec{r}_1 \cdot\left(\vec{m}_1+\lambda \vec{m}_2\right)=q_1+\lambda q_2, \lambda \in R \\
\Rightarrow & \vec{r} \cdot[(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+\hat{j}-2 \hat{k})]=5+\lambda(7) \\
\Rightarrow & \vec{r} \cdot[\hat{i}(1+3 \lambda)+\hat{j}(-2+\lambda)+\hat{k}(3-2 \lambda)]=5+7 \lambda
\end{aligned}
$$ ...(iii)
$$
\begin{aligned}
& \text { but given that } \vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k}) \\
& \Rightarrow 2(1+3 \lambda)+(-3)(-2+\lambda)+(1)(3-2 \lambda)=5+7 \lambda \\
& \Rightarrow 11+\lambda=5+7 \lambda \Rightarrow 6 \lambda=6 \Rightarrow \lambda=1 .
\end{aligned}
$$
putting $\lambda=1$ in equation (iii), we get
$$
\begin{aligned}
& \Rightarrow \quad \vec{r} \cdot[\hat{i}(1+3)+\hat{j}(-2+1)+\hat{k}(3-2)]=5+7 \\
& \Rightarrow \quad \vec{r} \cdot[4 \hat{i}-\hat{j}+\hat{k}]=12
\end{aligned}
$$
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