# Lecture - 8 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:

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

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

If intercepts on the axes are given then equation of plane in the Cartesian form:

Intercept on x-axis: A

Intercept on y-axis: B

Intercept on z-axis: C

We can use identity for three passsing points to find equation of plane.

Equation of plane: \(\frac{x}{A}+\frac{y}{B}+\frac{z}{C}=1\)

**6.** The foot of perpendicular drawn from origin to the plane is (4, −2, −5). Find the equation of the plane.

**17.** Write the equation of the plane whose intercepts on the coordinate axes are 2, -3 and 4.

**18.** Reduce the equations of the following planes in intercept form and find its intercepts on the coordinate axes: * a*. 4

*x*+3

*y*-6

*z*-12=0

*. 2*

**b***x*+3

*y*–

*z*=6

*. 2*

**c***x*–

*y*+

*z*=5

**19.** Find the equation of a plane which meets the axes in A, B and C, given that the centroid of the triangle ABC is the point \((\alpha,\beta,\gamma)\) .

**20.** Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes.

**21.** A plane meets the coordinate axes at A, B and C respectively such that the centroid of triangle ABC is (1, −2, 3). Find the equation of the plane.