find the distance from the point A(-2,1,2) to the plane 3x-2y+5z+1 =0

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

The point A(-2,1,2)

The plane 3x-2y+5z+1=0

To find a distance between a point and a plane:

 The standard plane formula is Ax+By+Cz+D =0

which in our case 3x-2y+5z+1=0

The standard point (x1,y1,z1) = (-2,1,2)

Then the distance d= l Ax1+By1+Cz1+D l/ sqrt(A^2+B^2+C^2)

Then in our case d = l 3(-2)+(-2)(1)+5(2)+1 l/ sqrt(9+4+25)

==> d= l -6-2+10+1 l/ sqrt(38) = 3/sqrt(38) = 3/6.16 = 0.487

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the length of perpendicular from A(-2,1,2) to the plane (2y+5z+1) = 0.

Solution:

The length of perpendicular p from a point P(x1,y1,z1) to the plane ax+by+cz+d = 0 is given by:

p = (ax1+by1+cz1+d)/(a^2+b^2+c^2)^(1/2). Therefore,

The length of perpendicular from A(-2,1,2) to the plane (3x-2y+5z+1) = 0. is given by:

p = [3(-2)-2(1)+5(2)]/(3^2+(-2)^2+5^2)^(1/2)

=3/38^(1/2)

 

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