How Do You Find the Equation of a Plane Given a Point and a Perpendicular Line?

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SUMMARY

The equation of a plane can be determined using a point and a direction vector from a line. Given the point (1, 1, -1) and the line defined by points (2, 0, 1) and (1, -1, 0), the direction vector is calculated as (1, 1, 1). The plane's equation takes the form ax + by + cz + d = 0, where the coefficients satisfy the conditions a + b + c = 0 and a + b - c + d = 0, incorporating the given point.

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  • Familiarity with dot product calculations
  • Basic algebra for solving linear equations
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salistoun
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Hi all,

How do you go about solving the following Question?

Find an equation of the plane containing the point (1 , 1 , -1) and perpendicular to the line through the points (2 , 0 , 1) and (1 , -1 , 0).

Stephen
 
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salistoun said:
Hi all,

How do you go about solving the following Question?

Find an equation of the plane containing the point (1 , 1 , -1) and perpendicular to the line through the points (2 , 0 , 1) and (1 , -1 , 0).

Stephen


Direction vector of the line:

$$\underline v:=(2,0,1)-(1,-1,0)=(1,1,1)$$

So plane perpendicular to the above line is

$$ax+by+cz+d=0\,\,,\,\,with\,\,\,0=(a,b,c)\cdot (1,1,1)=a+b+c=0$$

But this plane also has to contain the point [itex]\,(1,1,-1)\,[/itex] , so it has also to be

$$a+b-c+d=0$$

Now try to continue from here and end the exercise.

DonAntonio
 
Thanks Don it makes sense.

Stephen
 

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