Find the equation of the plane through the points A( 3, 4, 2) and B(7, 0, 6) and perpendicular to the plane 2x – 5y = 15.

Question

Find the equation of the plane through the points A( 3, 4, 2) and B(7, 0, 6) and perpendicular to the plane 2x – 5y = 15. HINT: The given plane is 2x – 5y + 0z = 15

Philoid · Accepted Answer

Plane passes through (3,4,2) and (7,0,6),

A(x-3) + B(y-4) + C(z-2)=0 (1)

A(x-7) + B(y-0) + C(z-6)=0 (2)

Subtracting (1) from (2),

A(x-7-x + 3) + B(y-y + 4) + C(z-6-z + 2)=0

-4A + 4B-4C=0

A-B + C=0

B=A + C (3)

Now plane is perpendicular to 2x-5y=15

2A-5B=0 (4)

Using (3) in (4)

2A-5(A + C)=0

2A-5A-5C=0

-3A-5C=0

Putting values in equation (1)

A(5(x-3) + 2(y-4)-3(z-2)=0

5x + 2y-3z-15-8 + 6=0

5x + 2y-3z-17=0

So, required equation of plane is 5x + 2y-3z-17=0.

Find the equation of the plane through the points A( 3, 4, 2) and B(7, 0, 6) and perpendicular to the plane 2x – 5y = 15.HINT: The given plane is 2x – 5y + 0z = 15

Find the equation of the plane through the points A( 3, 4, 2) and B(7, 0, 6) and perpendicular to the plane 2x – 5y = 15.
HINT: The given plane is 2x – 5y + 0z = 15