ONE MORE STEP TOWARD BETTER TOMMOROW
Saturday, 9 August 2014
CHAPTER 6 - Worked Out Examples
Find the equation of the plane passing through the points
be any arbitrary point in the plane whose equation we wish to determine:
will be perpendicular to this plane, we must have
We could have proceeded alternatively as follows: using the result of the last example, any arbitrary plane through
will be of the form
If this passes through
, we have
Thus the equation of the plane is
Find the equation of the plane intercepting lengths
The plane passes through the points
. For any variable point
in this plane, we have (as discussed in the previous section),
This general equation has the same form as the equation of the line in intercept form; which further proves the analogy between the formulae in two and in three dimensions.
A plane is at a distance
from the origin and the direction cosines of the (outward) normal to it are
. Find its equation.
The unit vector
normal to the plane is
For any point
in the plane, we have
This is the required equation; it is called the normal form of the plane’s equation. As an exercise, convert the general equation of the plane
into normal form.
August 09, 2014
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