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

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

Where, is the point on the plane and is normal vector.

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

Therefore, the equation of the plane is .

step-by-step explanation:

what is the figure? attach a picture.

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.