The shortest distance between the lines $x+y+2z-3=2x+3y+4z-4=0$ and the z-axis is?. $$\text {Attempt} $$. I first took the cross product of the vectors of two lines ie $i+j+2k,2i+3j+4k $ to get the perpendicular vector as $-2i+k $. I know that distance between two skew lines is $|\frac {(a_2-a_1). (b_1×b_2)}{(b_1×b_2)}| $. Where $a_2,a_1$ are position vectors from lines and $b_1,b_2$ are the vectors $-2i+k,k (z-axis) $.But how to find the position vectors ?.Thank you.
[Math] Shortest distance from z-axis to given lines
3dplane-geometrysolid-geometryvector-spacesvectors
Related Solutions
So I figured answer:
(iii) Using the distance formula, the shortest distance D from B to π1 is given by:
D = |ax + by + cz + d| / |n| where n is the normal to plane; a, b, c, d are the coefficients of the equation of the plane; x, y, z are the coordinates of the point from the plane.
π1: 8i + j - 14k = 75
B: 2i + 3j + 5k
D = |(8)(2) + (1)(3) + (-14)(5) + (-75)| / sqrt(64+1+196)
D = |-126| / sqrt(261)
D = 126/16.155 = 7.799 ~= 7.80
First the equations for all vectors $x$ on line $g$ and all vectors $y$ on line $h$: $$ \begin{align} g: x &= a + \lambda b \quad \\ h: y &= c + \mu d \end{align} \quad (*) $$ The difference vector between two of those vectors is $$ D = y - x = c - a + \mu d - \lambda b $$ The length of $D$ squared is: \begin{align} q(\lambda, \mu) &= D \cdot D \\ &= (c - a)^2 + (\mu d-\lambda b)^2 + 2 (c-a)\cdot(\mu d - \lambda b) \\ &= (c - a)^2 + \mu^2 d^2 + \lambda^2 b^2 - 2 \mu \lambda (d\cdot b) + 2 \mu ((c-a)\cdot d) - 2 \lambda ((c-a)\cdot b) \\ \end{align} The gradient of $q$ is: $$ q_\lambda = 2\lambda b^2 - 2\mu (d\cdot b)-2((c-a)\cdot b) \\ q_\mu = 2\mu d^2 - 2\lambda (d\cdot b)+2((c-a)\cdot d) $$ It should vanish for local extrema of $q$ which leads to the system $$ A u = v $$ with $$ A = \left( \begin{matrix} b^2 & - d\cdot b \\ - d \cdot b & d^2 \end{matrix} \right) \quad u = \left( \begin{matrix} \lambda \\ \mu \end{matrix} \right) \quad v = \left( \begin{matrix} (c-a) \cdot b \\ -(c-a) \cdot d \end{matrix} \right) $$ A unique solution exists for $$ 0 \ne \mbox{det A} = b^2 d^2 - (d \cdot b)^2 $$ Note that the dot operator stands for the scalar product. That solution is $$ u = A^{-1} v $$ with $$ A^{-1} = \frac{1}{\mbox{det } A} \left( \begin{matrix} d^2 & d\cdot b \\ d \cdot b & b^2 \end{matrix} \right) $$ Inserting the found values for $\lambda$ and $\mu$ into equations $(*)$ will provide you the two vectors, whose points are closest to each other.
Example:
$$ a = (2,0,0), b=(1,1,1), c=(0,1,-1), d=(-1,0,-1) $$ leads to $$ \lambda = 1, \mu = -2.5, x_\min = (3,1,1), y_\min=(2.5,1,1.5), D=(-0.5,0,0.5) $$
The image renders the $x$ values in green, the $y$ values in purple, and $D$ in red.
Best Answer
Let $z=t$.
Hence, $x+y=3-2t$ and $2x+3y=4-4t$, which gives $x=5-2t$ and $y=-2$.
Thus, $(5-2t,-2,t)$ is our line.
Now, let $(5-2t,-2,t)\subset\pi$ and $(0,0,k)||\pi$.
Let $\vec{n}(a,b,c)$ is a normal of $\pi$.
Thus, $(a,b,c)(0,0,1)=0$ and $(a,b,c)(-2,0,1)=0$ or $c=0$ and $-2a+c=0$,
which gives that $\vec{n}(0,1,0)$ and the equation of $\pi$ it's $y=-2$.
Id est, the distance is $2$.