Try this solution:
for the circle A: circumference=6π, area=9π
for the circle B: circumference=12π, area=36π

PS. formula for circumference is 'L=2πr', for area is 'S=πr²'.

8 0

3 years ago

find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).

maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).

To find:

The component equation of the plane.

Solution:

The equation of plane is

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Where, $(x_0,y_0,z_0)$ is the point on the plane and $\left< a,b,c\right>$ is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

$-2(x-(-10))+5(y-7)+1(z-5)=0$

$-2(x+10)+5y-35+z-5=0$

$-2x-20+5y-35+z-5=0$

$-2x+5y+z-60=0$

$-2x+5y+z=60$

Therefore, the equation of the plane is $-2x+5y+z=60$ .

4 0

2 years ago