Let t be the plane-2x-2y+z =-13. find the shortest distance d from the point po=(-5,-5,-3) to t, and the point q in t that is closest to po. use the square root symbol 'where needed to give an exact value for your answer. d 0 0 (0, 0, 0)

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane , then r will have the next parametric equations:

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

Substitute the value of in the parametric equations:

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

the answer is:

the correct option:

b.

to solve the problem, we need to remember how to operate to simplify exponential expressions using the logarithms.

we are given the equation:

then, using placing logarith with base 7 on both sides of the equation, we have:

so, rounding the obtained result to two decimal places (nearest hundreth), we have:

hence, we have that the correct option:

b.

have a nice day!