# Lecture - 6 Chapter 11 Three Dimensional Geometry

“Whenever you find yourself on the side of the majority, it is time to pause and reflect.” –Mark Twain

In this video, I am discussing equation of plane in three dimensional geometry and its derivation.

If One passing point and a normal vector is given then equation of plane in both the vector form and the Cartesian form:

\((\vec{r}-\vec{a})\cdot\vec{n}=0 \)

\((x-x_1)a+(y-y_1)b+(z-z_1)c=0\)

If three non-collinear point are given then equation of plane in both the vector form and the Cartesian form:

\((\vec{r}-\vec{a})\cdot[(\vec{b}-\vec{a})\times(\vec{c}-\vec{a})]=0\)

\(\left|\begin{matrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\\\end{matrix}\right|=0\)

1. Find the vector and the cartesian equations of the plane which passes through the point (5,2,-4) and perpendicular to the line with direction ratios 2, 3, −1.

3. Find the vector and the cartesian equations of the planes that passes through the point (1, 0, −2) and the normal to the plane is \(\hat{i}+\hat{j}-\hat{k}\).

4. Find the vector and the Cartesian equations of the plane that passes through points (1, 1, 0), (1, 2, 1), (−2, 2, −1).

5. Find the equation of the plane passes through the points (1,1,−1), (6,4,−5), (−4,−2, 3).