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Question: How do I find the point-normal form of the equation of the plane containing the point (-3,-4,3) and ...

How do I find the point-normal form of the equation of the plane containing the point (-3,-4,3) and perpendicular to (4,1,-2)?

Explanation

Solution

This type of problem is based on the concept of planes. First, we have to consider the point- normal equation of the plane, that is, a(x−x0)+b(y−y0)+c(z−z0)=0a\left( x-{{x}_{0}} \right)+b\left( y-{{y}_{0}} \right)+c\left( z-{{z}_{0}} \right)=0. Here the plane passes through (-3,-4,3) and hence x0=−3,y0=−4{{x}_{0}}=-3,{{y}_{0}}=-4 and z0=3{{z}_{0}}=3. Then we need to compare the obtained equation of the plane with the point perpendicular to the plane, that is, (4,1,-2). Let us substitute a=4, b=1 and c=-2 in the obtained equation. We should then make some necessary calculations to obtain the required solution.

Complete step-by-step answer:
According to the question, we are asked to find the equation of the plane in normal form.
We have been given the points (-3,-4,3) passing through the plane.
We first have to find the point-normal equation passing through points (-3,-4,3).
We know that the plane passing through the point (x0,y0,z0)\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right) is
a(x−x0)+b(y−y0)+c(z−z0)=0a\left( x-{{x}_{0}} \right)+b\left( y-{{y}_{0}} \right)+c\left( z-{{z}_{0}} \right)=0 -------(1)
Here, let us substitute x0=−3,y0=−4{{x}_{0}}=-3,{{y}_{0}}=-4 and z0=3{{z}_{0}}=3 in equation (1).
We get,
a(x−(−3))+b(y−(−4))+c(z−3)=0a\left( x-\left( -3 \right) \right)+b\left( y-\left( -4 \right) \right)+c\left( z-3 \right)=0
On further simplification, we get,
a(x+3)+b(y+4)+c(z−3)=0a\left( x+3 \right)+b\left( y+4 \right)+c\left( z-3 \right)=0 ---------(2)

Now we have to find the directional ratios a,b and c.
But we know that, for a point perpendicular to the plane, the point will be the directional ratios.
Therefore,
a=4, b=1 and c=-2
Substituting the obtained values of directional ratios in equation (2), we get
4(x+3)+1(y+4)+(−2)(z−3)=04\left( x+3 \right)+1\left( y+4 \right)+\left( -2 \right)\left( z-3 \right)=0
On further simplification, we get,
4(x+3)+(y+4)+(−2)(z−3)=04\left( x+3 \right)+\left( y+4 \right)+\left( -2 \right)\left( z-3 \right)=0
∴4(x+3)+(y+4)−2(z−3)=0\therefore 4\left( x+3 \right)+\left( y+4 \right)-2\left( z-3 \right)=0
Therefore, the point-normal equation is 4(x+3)+(y+4)−2(z−3)=04\left( x+3 \right)+\left( y+4 \right)-2\left( z-3 \right)=0.

Hence, the point-normal form of the equation of the plane containing the point (-3,-4,3) and perpendicular to (4,1,-2) is 4(x+3)+(y+4)−2(z−3)=04\left( x+3 \right)+\left( y+4 \right)-2\left( z-3 \right)=0.

Note: Whenever you get this type of problem, we should make sure about the formula of the normal equation. We should avoid calculation mistakes based on sign conventions. We should be thorough with the formulas of all forms of plane equation. It is always advisable to find the plane equation with the points first and then solve the rest of the part. We should not get confused with slope intercept form.