Which of these points is the closest to the point (7, 1) ?

Step-by-step explanation:

Solution

The general equation of the plane passing through (2,5,−3) is

a(x−2)+b(y−5)+c(z+3)=0 (1)

Now, this plane passes through B(−2,−3,5) and C(5,3,−3).Then we have

a(−2−2)+b(−3−5)+c(5+3)=0

⇒−4a−8b+8c=0

⇒−a−2b+2c=0 (2)

a(5−2)+b(3−5)+c(−3+3)=0

⇒3a−2b=0 (3)

Solving (2) and (3) we get

0+4

a

=

6−0

b

=

2+6

c

⇒

4

a

=

6

b

=

8

c

=λ(say)

⇒a=4λ,b=6λ,c=8λ

Substituting the values of a,b and c in (1), we get

4λ(x−2)+6λ(y−5)+8λ(z+3)=0

⇒4(x−2)+6(y−5)+8(z+3)=0

⇒4x+6y+8z−14=0

⇒2x+3y+4z−7=0

This is the equation of the plane determined by the points A(2,5,−3),B(−2,−3,5) and C(5,3,−3)

Now, the distance of this plane from the point (7,2,4) is

D=

2

2

+3

2

+4

2

∣(2×7+3×2+4×4)−7∣

=

4+9+16

∣(14+6+16)−7∣

=

4+9+16

∣36−7∣

=

29

29

=

29

29

×

29

29

=

29

units.

Thus, the distance between the point (7,2,4) and the plane 2x+3y+4z−7=0 is

29

units.

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.