If (2,-1,3) is the foot of the perpendicular drawn from the origin to a plane, then the equation of that plane is
2x + y - 3z + 6 = 0
2x - y + 3z -14 = 0
2x - y + 3z - 13 = 0
2z + y + 3z - 10 = 0
To solve the problem, we need to find the equation of a plane given the foot of the perpendicular from the origin to the plane.
1. Identify the Given Information:
Let the plane be denoted as $\pi$ and the origin as $O = (0, 0, 0)$.
The foot of the perpendicular from the origin to the plane is $P = (2, -1, 3)$.
2. Determine the Normal’s Direction Ratios:
The direction ratios of the normal to the plane are given by the coordinates of $P$, which are $(2, -1, 3)$.
3. Form the Plane Equation:
The equation of the plane is of the form $2x - y + 3z = d$, where $d$ is a constant.
Since $P = (2, -1, 3)$ lies on the plane, substitute its coordinates into the equation:
$2(2) - (-1) + 3(3) = d$
$4 + 1 + 9 = d$
$d = 14$.
4. Write the Final Equation:
The equation of the plane is $2x - y + 3z = 14$, or $2x - y + 3z - 14 = 0$.
Final Answer:
The equation of the plane is $2x - y + 3z - 14 = 0$.
Match the following:
The inverses of exponential functions are the logarithmic functions. The exponential function is y = ax and its inverse is x = ay. The logarithmic function y = logax is derived as the equivalent to the exponential equation x = ay. y = logax only under the following conditions: x = ay, (where, a > 0, and a≠1). In totality, it is called the logarithmic function with base a.
The domain of a logarithmic function is real numbers greater than 0, and the range is real numbers. The graph of y = logax is symmetrical to the graph of y = ax w.r.t. the line y = x. This relationship is true for any of the exponential functions and their inverse.
Exponential functions have the formation as:
f(x)=bx
where,
b = the base
x = the exponent (or power)