Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane is given in the form $\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$, where $\vec{a}=\hat{i}-\hat{j}$, $\vec{b}=\hat{i}+\hat{j}+\hat{k}$, and $\vec{c}=\hat{i}-2\hat{j}+3\hat{k}$.

The normal vector to the plane is $\vec{n}=\vec{b}\times \vec{c}$.

$\vec{n}=(\hat{i}+\hat{j}+\hat{k})\times (\hat{i}-2\hat{j}+3\hat{k})$

$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -2& 3\end{array}\right|$

$\vec{n}=\hat{i}(3-(-2))-\hat{j}(3-1)+\hat{k}(-2-1)=5\hat{i}-2\hat{j}-3\hat{k}$

The equation of the plane is $\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$.

$\vec{a}\cdot \vec{n}=(\hat{i}-\hat{j})\cdot (5\hat{i}-2\hat{j}-3\hat{k})=1(5)+(-1)(-2)+0(-3)=5+2=7$

The equation of the plane in Cartesian form is $5x-2y-3z-7=0$.

The perpendicular distance of the plane from the origin $(0,0,0)$ is given by:

$d=\frac{|-7|}{\sqrt{5^2+(-2{)}^2+(-3{)}^2}}=\frac{7}{\sqrt{25+4+9}}=\frac{7}{\sqrt{38}}$ units.

Answer: $\frac{7}{\sqrt{38}}$ units