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Question: How do you find the equation of the plane in XYZ-space through the point \(p=\left( 4,5,4 \right)\) ...

How do you find the equation of the plane in XYZ-space through the point p=(4,5,4)p=\left( 4,5,4 \right) and perpendicular to the vector n=(−5,−3,−4)n=\left( -5,-3,-4 \right)?

Explanation

Solution

In the question given that we have to find the equation of a plane passing through the point p=(4,5,4)p=\left( 4,5,4 \right) and perpendicular to the vector n=(−5,−3,−4)n=\left( -5,-3,-4 \right). As we know that the if a plane is passing through the point p(x1,y1,z1)p\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and perpendicular to a vector v→ (a,b,c)\overset{\to }{\mathop{v}}\,\left( a,b,c \right) then the equation of the plane is written as a(x−x1)+b(y−y1)+c(z−z1)=0a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0. By using this formula, we will get the equation of the plane.

Complete step by step answer:
From the above question we have to find the equation of the plane,
From the question given that a plane passing through the point
⇒p=(4,5,4)\Rightarrow p=\left( 4,5,4 \right)
And also given that plane is perpendicular to the vector,
⇒n=(−5,−3,−4)\Rightarrow n=\left( -5,-3,-4 \right)
As we know that the if a plane is passing through the point p(x1,y1,z1)p\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and perpendicular to a vector v→ (a,b,c)\overset{\to }{\mathop{v}}\,\left( a,b,c \right) then the equation of the plane is written as
⇒a(x−x1)+b(y−y1)+c(z−z1)=0\Rightarrow a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0
By comparing with general formula, we will get the values as
⇒x1=4y1=5 z1=4\Rightarrow {{x}_{1}}=4\quad {{y}_{1}}=5\ \quad {{z}_{1}}=4
⇒a=−5b=−3c=−4\Rightarrow a=-5\quad b=-3\quad c=-4
Now we have to substitute the values in their respective position in the formula,
By substituting the values, we will get,
⇒(−5)(x−4)+(−3)(y−5)+(−4)(z−4)=0\Rightarrow \left( -5 \right)\left( x-4 \right)+\left( -3 \right)\left( y-5 \right)+\left( -4 \right)\left( z-4 \right)=0
By further simplification we will get,
⇒−5x+20−3y+15−4z+16=0\Rightarrow -5x+20-3y+15-4z+16=0
By further simplification we will get,
⇒5x+3y+4z−51=0\Rightarrow 5x+3y+4z-51=0
Therefore, the equation of the plane is ⇒5x+3y+4z−51=0\Rightarrow 5x+3y+4z-51=0

Note: Students should know the general formulas and relations of the Cartesian coordinate system, if students are given direction ratios of planes instead of perpendicular to the vector the formula is the same, students should not be confused.