Question

please help me in solving this problem of 3D??

Answer 1

find the distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to the line x/2=y/3=z/-6

Answer 2

the normal of the plane is: <1,-1,1>

apply this to the point: (1,-2,3)

x = 1+t
y = -2-t
z = 3+t

any point on our plane satisfies:

x-y+z=5, so lets sub this in to find "t" along the normal

1+t-(-2-t)+3+t = 5

1+t+2+t+3+t = 5

6+3t = 5

t = -1/3; this should be the distance the point is hovering away from the plane itself, a perp measure.

Answer 3

x = 1 -1/3 = 2/3
y = -2+1/3 = -5/3
z = 3- 1/3 = 8/3


the point on the plane directly beneath directly beneath the given should be:

(2/3 ,-5/3, 8/3); which turns out good.

the distance from this point to our hovering point is then:

3/3, -6/3, 9/3
-(2/3 ,-5/3, 8/3)
---------------
1/3, 11/3, 17/3

sqrt(1/9 + 121/9 + 289/9)

sqrt(411)/3 above the plane; but thats just side track stuff. we can use the same concept to find the distance along the other line given

Answer 4

x/2=y/3=z/-6

t=x/2
t=y/3
t=z/-6

2t+0= x
3t+0= y
-6t+0=z

we have the vector part, now apply our given point

2t+1= x
3t-2= y
-6t+3=z

and find where it hits the plane

2t+1 -3t+2 -6t+3 = 5

-7t+6 = 5

t = 1/7

2(1/7)+1 = x = 9/7
3(1/7)-2 = y = -11/7
-6(1/7)+3=z = 15/7

now we just need to measure the length between the given point and the plane point by the vector between them

Answer 5

7/7, -14/7, 21/7
-(9/7,-11/7, 15/7)
-----------------
<-2/7, -3/7, 6/7>

sqrt(4+9+36)/7

sqrt(49)/7 = 1

with any luck, that works :)