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.