Question:
The equation of the plane which contains the $\mathrm{y}$-axis and passes through the point $(1,2,3)$ is :
Correct Option: , 4
Solution:
$\overrightarrow{\mathrm{n}}=\hat{\mathrm{j}} \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
$=-3 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
So, $(-3)(x-1)+0(y-2)+(1)(z-3)=0$
$\Rightarrow-3 x+z=0$
Option 4
Alternate :
Required plane is
$\left|\begin{array}{lll}x & y & z \\ 0 & 1 & 0 \\ 1 & 2 & 3\end{array}\right|=0$
$\Rightarrow 3 x-z=0$