Question

An implicit equation for the plane passing through the points (1,3,−4), (−3,2,−7), and (1,0,−3) is

Answer 1

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Answer 2

Given the points $(1,3,-4),(-3,2,-7),(1,0,-3)$.

Need to find a plane equation that passes through the given points.

• For this first find two vectors as follows:

$QP=(1,3,-4)-(-3,2,-7)\\QP=(1-(-3),3-2,-4-(-7))\\QP=(4,1,3)$

$QR=(1,0,-3)-(-3,2,-7)\\QR=(1-(-3),0-2,-3-(-7))\\QR=(4,-2,4)$

• Now find the normal vector to the plane:

For this, find the cross product of $QP,QR$.

$QP$×$QR$=$\left[\begin{array}{ccc}i&j&k\\4&1&3\\4&-2&4\end{array}\right]$

$QP$×$QR$=$i(4+6)-j(16-12)+k(-8-4)$

$QP$×$QR$=$10i-4j-12k$

So, the normal to the plane is $(10,-4,-12)$.

• Now find the plane equation.

Definition: The plane equation that passes through the point $(x_{1} ,x_{2},x_{3} )$  with normal $(a,b,c)$ is $(a,b,c).(x-x_{1},y-y_{2},z-z_{3} )=0$.

Here, $(x_{1} ,x_{2},x_{3} )=(1,3,-4)$and $(a,b,c)=(10,-4,-12)$.

So,

 $(a,b,c).(x-x_{1},y-y_{2},z-z_{3} )=0\\(10,-4,-12).(x-1,y-3,z-(-4))=0\\10(x-1)-4(y-3)-12(z+4)=0\\10x-10-4y+12-12z-48=0\\10x-4y-12z-46=0\\10x-4y-12z=46$

Hence, the required plane equation in implicity form is $10x-4y-12z=46$.

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