Distance from $A(1, 2, 2)$ to the line $\frac{x-1}{-5}=\frac{y+1}{3}=\frac{z+1}{3}$

Question

I need to find the distance from the point $A(1, 2, 2)$ to the line $l:\frac{x-1}{-5}=\frac{y+1}{3}=\frac{z+1}{3}$.
I thought that I should find the perpendicular from $A$ to my line, intersect these two lines and then I have my distance.
Let's say that the perpendicular from $A$ to $l$ has the direction $(a, b, c)$. Then we must have that $-5a+3b+3c=0$.
Now, what I did was I chose the particular solution $(0, -1, 1)$ and I obtained the line $\frac{x-1}{0}=\frac{y-2}{-1}=\frac{z-2}{1}$, I intersected $l$ and this line and I got the distance to be $3\sqrt{2}$.
I would like to know if this is correct. I wonder if it is alright to pick a particular solution of $-5a+3b+3c=0$. I have a feeling that if this were right, I would be getting that there are multiple perpendiculars from $A$ to $l$ and this is wrong. Are all these lines the same?

Answer 1

No you cannot pick a particular solution to $a, b, c$. Please note that the equation of the line is,

$\displaystyle \small l: \frac{x-1}{-5}=\frac{y+1}{3}=\frac{z+1}{3}$, so a point $P$ on the line will be given by $x = 1-5t, y = z = 3t-1$

Hence the direction vector from point $A (1, 2, 2)$ to point $P$ on the line is $\vec {AP} = (-5t, 3t-3, 3t-3)$.

When $\vec {AP} $ is perpendicular to line $l$, the dot product of $\vec {AP}$ with the direction vector of line $l$ will be zero.

$-5 \cdot (-5t) + 3 \cdot (3t-3) + 3 \cdot (3t-3) = 0 \implies t = \frac{18}{43}$

Now plug in value of $t$ to find $|\vec{AP}|$ which is the distance you seek.

Answer 2

Line: $\frac{x-x_1}l=\frac{y-y_1}m=\frac{z-z_1}n$

Point:$A(x_0, y_0, z_0)$

$d=\frac{\sqrt{{\begin{vmatrix}y_0-y_1&z_0-z_1\\m&n\end{vmatrix}}^2+{\begin{vmatrix}z_0-z_1&x_0-x_1\\n&l\end{vmatrix}}^2+{\begin{vmatrix}x_0-x_1&y_0-y_1\\l&m\end{vmatrix}}^2}}{\sqrt{l^2+m^2+n^2}}$

Line:: $\frac{x-1}{-5}=\frac{y+1}3=\frac{z+1}3\rightarrow l=-5, m=3, n=3$

Point: $A(x_0=1,y_0= 2, z_0= 2)$

Plugging values in formula we get:

$d=\frac{15\sqrt{86}}{43}\approx 3.23$

Answer 3

The line $\frac{x−1}{−5}=\frac{y+1}{3}= \frac{z+1}{3}$ has tangent vector <-1/5, 1/3, 1/3> which is, of course, a normal vector to any plane perpendicar to that line. In particular, the plane containing the point (1, 2, 2) perpendicular to that line is (-1/5)(x- 1)+ (1/3)(y- 2)+ (1/3(z- 2)= 0 or -3(x- 1)+ 5(y- 2)+ 5(z- 3)= -3x+ 5y+ 5z- 22= 0 so -3x+ 5y+ 5z= 22.

The line $\frac{x−1}{−5}=\frac{y+1}{3}= \frac{z+1}{3}$ has parametic equation x= -5t+ 1, y= 3t- 1, z= 3t- 1. That line will intersect the plane where $-3(-5t+ 1)+ 5(3t- 1)+ 5(3t- 1)= 45t- 7= 22. 45t= 29 so t= 29/45. That is, the point of intersectoin is x= -5(29/45)+ 1= -29/9+ 1= -20/9, y= 3(29/45)- 1= 29/15- 1= 14/15, and z= 3(29/45)- 1= 29/15- 1= 14/15.

All that is left is to compute the distance between (1, 2, 2) and (-20/9, 14/15, 14/15)

Answer 4

After converting to parametric form, $(x,y,z) = (1-5t, 3t-1, 3t-1)$.

Thus for any $t$, the squared distance to the point $(1,2,2)$ is: $$(1-5t-1)^2 + 2(3t-1-2)^2 = 25t^2 + 2(9t^2 - 18t + 9) = 43t^2 - 36t +18.$$

Since $a > 0$, this quadratic is concave up, and hence the minimum distance is attained at $t = -\frac{b}{2a} = -\frac{-36}{2 (43)} = \frac{18}{43}$. Substituting this value of $t$ in gives $d = \sqrt{\frac{450}{43}} \approx 3.23$.

Answer 5

The parametric equation for the line gives Po (1,-1,-1) and its direction vector as -5i+3j+3k. First method: Take the dot product of Po>A with the direction vector and use this to find the angle in between. The perpendicular distance from A to the line is then the modulus of Po>A times the sine of that angle. ... Po>A = 0i+3j+3k and so the dot product of Po>A is 18. The modulus of Po>A is Sqrt(18) and the modulus of the direction vector is Sqrt(43), and so the angle is the inverse cosine of 18/(sqrt(18)*sqrt(43)) = 49.68 deg. The distance is sqrt(18)*sin(49.68) = 3.23. Second method: Since we see that the distance involves the sine of the angle, we could use the cross product between Po>A and the direction vector, taking the modulus of this product and dividing by the modulus of the direction vector only. The cross product is 0i+15j+15k, and the modulus of this is sqrt(450). The distance is given by sqrt(450)/sqrt(43) = 3.23. Done.