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Question: How to find the linear equation of the plane through the point \(\left( {1,2,3} \right)\) and contai...

How to find the linear equation of the plane through the point (1,2,3)\left( {1,2,3} \right) and contains the line represented by the vector equation
r(t)=3t,62t,12tr\left( t \right) = \left\langle {3t,6 - 2t,1 - 2t} \right\rangle ?

Explanation

Solution

Here, we are required to find the linear equation of the plane through the given point which contains the line with the given vector equation. Thus, we will use the parametric form of equation of line and substitute the given values and with the help of plane and vector equation, we will solve it further to find the required linear equation of the given plane.

Formula Used:
1. The parametric line equation is given by: p=p0+tvp = {p_0} + t \cdot \overrightarrow v
2. The plane equation is given by: w,pp1=0\left\langle {\overrightarrow w ,p - {p_1}} \right\rangle = 0

Complete step-by-step answer:
As we know,
The parametric line equation is given by:
p=p0+tvp = {p_0} + t \cdot \overrightarrow v
Where, p = \left\\{ {x,y,z} \right\\}, {p_0} = \left\\{ {0,6,1} \right\\} and \overrightarrow v = \left\\{ {3, - 2, - 2} \right\\}
Also, the plane equation is given by
w,pp1=0\left\langle {\overrightarrow w ,p - {p_1}} \right\rangle = 0
Here, according to the question, we have, {p_1} = \left\\{ {1,2,3} \right\\}and w\overrightarrow w is perpendicular to v\overrightarrow v and to the segment p1p0{p_1} - {p_0} such that:
\overrightarrow w = \overrightarrow v \times \left( {{p_1} - {p_0}} \right) = \left\\{ {3, - 2, - 2} \right\\} \times \left\\{ {1, - 4, - 2} \right\\}
Then,
\overrightarrow w = \left\\{ { - 4,4, - 10} \right\\}
The plane equation is given as:
(4)(x1)+4(y2)+(10)(z3)=0\left( { - 4} \right)\left( {x - 1} \right) + 4\left( {y - 2} \right) + \left( { - 10} \right)\left( {z - 3} \right) = 0
4x+4+4y810z+30=0\Rightarrow - 4x + 4 + 4y - 8 - 10z + 30 = 0
Solving the like terms further, we get,
4x+4y10z+26=0\Rightarrow - 4x + 4y - 10z + 26 = 0

Therefore, the linear equation of the plane through the point (1,2,3)\left( {1,2,3} \right) which contains the line represented by the vector equation r(t)=3t,62t,12tr\left( t \right) = \left\langle {3t,6 - 2t,1 - 2t} \right\rangle is represented by 4x+4y10z+26=0 - 4x + 4y - 10z + 26 = 0.
Hence, this is the required answer.

Note:
Parametric equation is a type of equation that employs an independent variable called a parameter (often denoted by tt) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable. Also, a vector equation is any function that takes any one or more variables and returns a vector. The vector equation of a line is an equation that identifies the position vector of every point along the line.