Distance from P to Line: Find Q Closest to P

[Kaede_N9]

Homework Statement

In each case find the shortest distance from the point P to the line, and find the point Q on the line closest to P.

P(-1,0,1); [x y z]^T = [3 1 -4]^T +t[3 -2 0]

Homework Equations

Projection equation

The Attempt at a Solution

Let P_o be (3,1-4)

Vector V = Vector P_o to P = (-1,0,1) - (3,-1,4) = (-4,-1,-3)

Vector V_1 = projection of vector V over direction vector
= ( ( (-4,-1,-3)·(3,-2,0) ) / (sqrt. ((3)^2 + (-2)^2)) (3,-2,0)
= (-10 / 13) (3,-2,0)

||Vector QP|| = || vector V - vector V_1 ||
= || (-4,-1,-3) + (30/13,-20/13,0)||
=|| (-52/13,-13/13,-3) + (30/13,-20/13,0)||

= ||(-22/13,-33/13,-3/13)||
= (1/13) (sqrt.1582)

In the back of the book it says (1/13) (sqrt.1846). Where did I go wrong?

 

[HallsofIvy]

Homework Statement

In each case find the shortest distance from the point P to the line, and find the point Q on the line closest to P.

P(-1,0,1); [x y z]^T = [3 1 -4]^T +t[3 -2 0]

Homework Equations

Projection equation

The Attempt at a Solution

Let P_o be (3,1-4)

Vector V = Vector P_o to P = (-1,0,1) - (3,-1,4) = (-4,-1,-3)

How did "(3, 1, -4)" before become (3, -1, 4)? Was that a typo? Unfortunately, that error propogates. The vector from P_0 to P is (-1, 0, 1)- (3, 1, -4)= (-4, -1, 5), not (-4, -1, -3).

Vector V_1 = projection of vector V over direction vector
= ( ( (-4,-1,-3)·(3,-2,0) ) / (sqrt. ((3)^2 + (-2)^2)) (3,-2,0)
= (-10 / 13) (3,-2,0)

Vector QP = || vector V - vector V_1 ||
= || (-4,-1,-3) + (30/13,-20/13,0)||
=|| (-52/13,-13/13,-3) + (30/13,-20/13,0)||

= ||(-22/13,-33/13,-3/13)||
= (1/13) (sqrt.1582)

In the back of the book it says (1/13) (sqrt.1846). Where did I go wrong?

Personally, I would have done this in a completely different way: the plane through (-1, 0, 1) perpendicular to the given line is 3(x+ 1)- 2(y- 0)+ 0(z- 1)= 3x+ 3- 2y= 0 or 3x- 2y= -3. The line, x= 3+ 3t, y= 1- 2t, z= 4, crosses that plane when 3(3+ 3t)- 2(1-2t)= 9+ 9t- 2+ 4t= 13t+ 7= -3 so that t -10/13. That is, x= 3+ 3(-10/13= (39- 30)/13= 9/13. y= 1- 2(9/13)= (13- 18)/13= -5/13, z= 4. Find the distance from (-1, 0, 1) to (9/13, -5/13, 4).

 

[Kaede_N9]

How did "(3, 1, -4)" before become (3, -1, 4)? Was that a typo? Unfortunately, that error propogates. The vector from P_0 to P is (-1, 0, 1)- (3, 1, -4)= (-4, -1, 5), not (-4, -1, -3).

Yes, it was a typo. The point should be (3, -1, 4) , and not (3, 1, -4).

Personally, I would have done this in a completely different way: the plane through (-1, 0, 1) perpendicular to the given line is 3(x+ 1)- 2(y- 0)+ 0(z- 1)= 3x+ 3- 2y= 0 or 3x- 2y= -3. The line, x= 3+ 3t, y= 1- 2t, z= 4, crosses that plane when 3(3+ 3t)- 2(1-2t)= 9+ 9t- 2+ 4t= 13t+ 7= -3 so that t -10/13. That is, x= 3+ 3(-10/13= (39- 30)/13= 9/13. y= 1- 2(9/13)= (13- 18)/13= -5/13, z= 4. Find the distance from (-1, 0, 1) to (9/13, -5/13, 4).

For the the line, x= 3+ 3t, y= 1- 2t, z= 4 , should 1 or 4, in y and z respectively, be negative? (since the point, at the time you typed your response, could have been (3, -1, 4) or (3, 1, -4).

Thank you for your help. I will try your method and see if I get the correct answer.

 

[Kaede_N9]

Using your method I almost obtained the correct answer.

P(-1,0,1); [x y z]^T = [3 -1 4]^T +t[3 -2 0]

3(x-1) - 2(y-0) - 0(z-1)
= 3x -3-2y
or
-3 = 3x -2

and
x = 3+3t
y = -1-2t
z = 4

---

-3 = 3(3 +3t) - 2(-1-2t)
-3 = 11 + 13t
-14/13 = t

---

x = 3 - 42/13 = -3/13
y = -1 + 28/13 = 15/13
z = 4

(-3/13, 15/13, 4) <------- Point Q
---

(-1,0,1) - (-3/13, 15/13, 4)
= (10/13, 15, 3)

||(10/13, 15, 3)||
=(1/13)(sqrt1846) <------- Distance from Point Q to Point P

--
So I obtained the correct magnitude for QP and point for Q just like in the back of the book.

Thank you very much for your assistance!