Question

find the distance of pt p (1,2,3) to xy plaaane

Answer 1

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Let's find equation of straight line passing through (1,-2,3) and parallel to x/2 = y/3 = z/-6

As direction ratio would be 2,3,-6

So eq of line is x-1)/2 = y+2)/3 = z-3)/-6

Let any point on this line be 2a+1, 3a-2,-6a+3

Let's find intersection with 4x -y +z= 5

4( 2a+1) -(3a-2) -6a +3 = 5

8a + 4 -3a +2 -6a +3 = 5

-a +4 = 0

a= 4

so point is 2(4) +1, 3(4) -2, -6(4) +3

= 9, 10,-21

Distance b/w (1,-2,3) and (9,10,-21)

√(9–1)^2 + ( 10 +2)^2 + ( -21–3)^2

√8^2 + 12^2 + 24^2

√64 +144 + 576

√784

28

so distance is 28.

Answer 2

Let's find equation of straight line passing through (1,-2,3) and parallel to x/2 = y/3 = z/-6

As direction ratio would be 2,3,-6

So eq of line is x-1)/2 = y+2)/3 = z-3)/-6

Let any point on this line be 2a+1, 3a-2,-6a+3

Let's find intersection with 4x -y +z= 5

4( 2a+1) -(3a-2) -6a +3 = 5

8a + 4 -3a +2 -6a +3 = 5

-a +4 = 0

a= 4

so point is 2(4) +1, 3(4) -2, -6(4) +3

= 9, 10,-21

Distance b/w (1,-2,3) and (9,10,-21)

√(9–1)^2 + ( 10 +2)^2 + ( -21–3)^2

√8^2 + 12^2 + 24^2

√64 +144 + 576

√784

28

so distance is 28.