Difference between revisions of "Science:Math Exam Resources/Courses/MATH105/April 2017/Question 01 (a)/Solution 1"

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If <math> \langle x, y, z\rangle </math> is some point on the plane, then it forms a vector  <math> \bold{v}=( \langle x, y, z\rangle- \langle 0, -3,2 \rangle)</math>  with the given point on the plane <math> \langle 0, -3,2 \rangle </math>. We know that this vector  <math> \bold{v}</math> must be perpendicular to the normal vector  <math> \langle  2, -1 , 4 \rangle </math> of the plane.  
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If <math> \langle x, y, z\rangle </math> is some point on the plane, then it forms a vector  <math> \bold{v}=( \langle x, y, z\rangle- \langle 0, -3,2 \rangle)</math>  with the given point on the plane <math> \langle 0, -3,2 \rangle </math>. Since a normal vector on a plane is orthogonal to any vector lying on the plane, this vector  <math> \bold{v}</math> is therefore perpendicular to the normal vector  <math> \langle  2, -1 , 4 \rangle </math> of the plane.  
  
  

Latest revision as of 21:47, 8 March 2018

If is some point on the plane, then it forms a vector with the given point on the plane . Since a normal vector on a plane is orthogonal to any vector lying on the plane, this vector is therefore perpendicular to the normal vector of the plane.


This implies that .


Thus the equation of the plane is .