Find the distance of the point P(-2, 3, -4) from the line (x + 2)/3 =

1.	Find the distance of the point P(-2, 3, -4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y - 3z + 1 = 0.
Answer» Step 1: Equations of plane through P(-2, 3, -4) and parallel to the given plane is: 4(x + 2) + 12(y - 3) -3(z + 4) = 0 4x + 12y - 3z - 40 = 0 Step: General point on the line A(3r - 2, (4r -3)/2, (5r-4)/3) Step 3: A is on the given plane therefore 12r - 8 + 24r - 18 - 5r + 4 - 40 = 0 31r = 62 r = 62/31 r = 2 A(4, 5/2,2) Hence PA = √36 + 1/4 + 36 = √(144+ 1 + 144)/4 = √289/4 = 17/2

Find the distance of the point P(-2, 3, -4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y - 3z + 1 = 0.

Answer»

Step 1: Equations of plane through P(-2, 3, -4) and parallel to the given plane is:

4(x + 2) + 12(y - 3) -3(z + 4) = 0

4x + 12y - 3z - 40 = 0

Step: General point on the line

A(3r - 2, (4r -3)/2, (5r-4)/3)

Step 3: A is on the given plane therefore

12r - 8 + 24r - 18 - 5r + 4 - 40 = 0

31r = 62

r = 62/31

r = 2

A(4, 5/2,2)

Hence PA = √36 + 1/4 + 36

= √(144+ 1 + 144)/4

= √289/4

= 17/2

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Find the distance of the point P(-2, 3, -4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y - 3z + 1 = 0.

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