Plane
Equation of a Plane in Normal Form

Two lines, , are coplanar, if and only if

Two lines, ,
are coplanar, if
Solved ExamplesExample 1:
Show that the lines are coplanar.
Solution:
The given lines can be rewritten as
Comparing with , we obtain
Hence, the given lines are coplanar.
Example 2:
For what value of p are the following two lines coplanar?
Solution:
Comparing the given lines with, we obtain
$\left(\overrightarrow{{a}_{2}}\overrightarrow{{a}_{1}}\right).\left(\overrightarrow{{b}_{1}}\times \overrightarrow{{b}_{2}}\right)=\left(2\hat{i}+5\hat{j}\hat{k}\right).\left(2p\hat{i}+5\hat{j}+p\hat{k}\right)=4p+25p=5p+25$
If these two lines are coplanar, then
Thus, the required value of p is 5.

Two lines, , are coplanar, if and only if

Two lines, ,
are coplanar, if
Solved ExamplesExample 1:
Show that the lines are coplanar.
Solution:
The given lines can be rewritten as
Comparing with , we obtain
Hence, the given lines are coplanar.
Example 2:
For what value of p are the following two lines coplanar?
Solution:
Comparing the given lines with, we obtain
$\left(\overrightarrow{{a}_{2}}\overrightarrow{{a}_{1}}\right).\left(\overrightarrow{{b}_{1}}\times \overrightarrow{{b}_{2}}\right)=\left(2\hat{i}+5\hat{j}\hat{k}\right).\left(2p\hat{i}+5\hat{j}+p\hat{k}\right)=4p+25p=5p+25$
If these two lines are coplanar, then
Thus, the required value of p is 5.

If a plane is given whose perpendicular distance from origin is d, and is the unit normal vector along the direction of the normal to the plane from the origin, then:

General vector equation of such a plane is given by

Equation of the plane when direction cosines of the unit normal vector are given as l, m, and n is lx + my + nz = d


Equation of the plane when the direction ratios of the unit normal vector are given as a, b, and c instead of direction cosines is:
Vector form− ; Cartesian form− ax + by + cz = d

The normal drawn from origin to plane lx + my + nz = d will intersect the plane at (ld, md, nd).

The normal drawn from origin to plane ax + bx + cz = d will intersect the plane at .
Example 1:
What is the equation of the plane which is at the distance of units from the origin and its normal vector from the origin is?
Solution:
Normal vector,
Hence, the equation of plane is
Example 2:
Find the distance of plane 3x − 4y + 22 − 9 = 0 from the origin.
Solution:
3x − 4y + 2z = 9
Here, it can be noticed that
(3)^{2} + (− 4)^{2} + (2)^{2} = 29 ≠ 1
Hence, equation is in the form of ax + by + cz = d
Dividing both sides by, we obtain
Distance of planeā¦
To view the complete topic, please