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If $\alpha$ and $\beta$ are scalar and $\mathbf{r}=(2+\alpha-3 \beta) \hat{\mathbf{i}}+(\beta-3) \hat{\mathbf{j}}+(2 \alpha-5 \beta-1) \hat{\mathbf{k}}$ is equation of a plane, then that equation in Cartesian form is
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Verified Answer
The correct answer is:
$2 x+y-z=2$
$\mathbf{r}=(2+\alpha-3 \beta) \hat{\mathbf{i}}+(\beta-3) \hat{\mathbf{j}}+(2 \alpha-5 \beta-1) \hat{\mathbf{k}}$
Let $\quad \mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
From Eq. (i), we get
$\begin{aligned}
x & =2+\alpha-3 \beta \\
y & =\beta-3 \Rightarrow \beta=y+3 \\
z & =2 \alpha-5 \beta-1 \Rightarrow z=2 \alpha-5(y+3)-1 \\
\Rightarrow z+5 y+16 & =2 \alpha \\
\Rightarrow \quad \alpha & =\frac{z+5 y+16}{2}
\end{aligned}$
From Eqs. (ii) and (iii), we get
$\begin{aligned}
x & =2+\left(\frac{z+5 y+16}{2}\right)-3(y+3) \\
\Rightarrow \quad 2 x & =4+z+5 y+16-6 y-18 \\
\Rightarrow \quad 2 x+y-z & =2
\end{aligned}$
Let $\quad \mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
From Eq. (i), we get
$\begin{aligned}
x & =2+\alpha-3 \beta \\
y & =\beta-3 \Rightarrow \beta=y+3 \\
z & =2 \alpha-5 \beta-1 \Rightarrow z=2 \alpha-5(y+3)-1 \\
\Rightarrow z+5 y+16 & =2 \alpha \\
\Rightarrow \quad \alpha & =\frac{z+5 y+16}{2}
\end{aligned}$
From Eqs. (ii) and (iii), we get
$\begin{aligned}
x & =2+\left(\frac{z+5 y+16}{2}\right)-3(y+3) \\
\Rightarrow \quad 2 x & =4+z+5 y+16-6 y-18 \\
\Rightarrow \quad 2 x+y-z & =2
\end{aligned}$
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