Line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$

Question

A straight line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$,then square of the perpendicular distance of origin from L is ?

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Let $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ be the required line.
Then $2l+3m-10n=0,4l-3m-2n=0$

So $l:m:n=1:1:\frac{1}{2}$

So the line is $\frac{x-x_1}{1}=\frac{y-y_1}{1}=\frac{z-z_1}{\frac{1}{2}}$.
But i could not solve further.Please help me.

Answer 1

Hint.

I give you a hint that you can apply to find the distance you're looking for.

So you have two lines $$\begin{cases} L_1 \equiv \frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10} \\ L_2 \equiv \frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}\end{cases}$$

For $L_1$, you can find a point $A_1 \in L_1$, for example $A_1=(-2,-6,34)$ and a direction vector of $L_1$, for example $u_1=(2,3,-10)$. You can do the same for $L_2$ to get $A_2$ and $u_2$.

Now the cross product $u = u_1 \times u_2$ is perpendicular to $u_1$ and $u_2$. The plane $P_1$ containing the line $L_1$ and the direction $u$ intersects $L_2$ at $B_2$. Similarly, The plane $P_2$ containing the line $L_2$ and the direction $u$ intersects $L_1$ at $B_1$. The line $B_1 B_2$ is the perpendicular to your orginal two lines.

Finally you can find its distance to the origin.

Answer 2

Let $L_1$ be $x=2s-2,y=3s-6,z=-10s+34$ and $L_2$ be $x=4t-6,y=-3t+7,z=-2t+7$.

As you wrote, $L$ can be written as $x=u+x_1,y=u+y_1,z=\frac 12u+z_1$.

Now, if we let $A(a,b,c)$ be a point where $L_1$ and $L$ intersect, then we have $$a=2s_1-2,b=3s_1-6,c=-10s_1+34,a=u_1+x_1,b=u_1+y_1,c=\frac 12u_1+z_1.$$ So, $$2s_1-2=u_1+x_1,3s_1-6=u_1+y_1,-10s_1+34=\frac 12u_1+z_1.$$ So, $$(u_1=)\ 2s_1-2-x_1=3s_1-6-y_1=2(-10s_1+34-z_1)$$ So, $$y_1=x_1+s_1-4,z_1=\frac 12x_1-11s_1+35\tag1$$

Now, if we let $B(d,e,f)$ be a point where $L_2$ and $L$ intersect, then we have $$d=4t_1-6,e=-3t_1+7,f=-2t_1+7,d=u_2+x_1,e=u_2+y_1,f=\frac 12u_2+z_1.$$ So, $$4t_1-6=u_2+x_1,-3t_1+7=u_2+y_1,-2t_1+7=\frac 12u_2+z_1.$$ So, $$(u_2=)\ 4t_1-6-x_1=-3t_1+7-y_1=2(-2t_1+7-z_1).$$ So, $$y_1=x_1-7t_1+13,z_1=\frac 12x_1-4t_1+10\tag2$$

From $(1)(2)$, solving $$(y_1=)\ x_1+s_1-4=x_1-7t_1+13\iff s_1-4=-7t_1+13$$ and $$(z_1=)\ \frac 12x_1-11s_1+35=\frac 12x_1-4t_1+10\iff -11s_1+35=-4t_1+10$$ gives $$s_1=3,t_1=2.$$

So, from $(1)$, we have $$y_1=x_1-1,z_1=\frac 12x_1+2.$$

Then, solving $x_1\times 1+(x_1-1)\times 1+\left(\frac 12x_1+2\right)\cdot \frac 12=0$ gives $x_1=0$.

Thus, the square of the perpendicular distance of origin from L equals the square of the distance of origin from a point $(0,-1,2)$, i.e. $$0^2+(-1)^2+2^2=\color{red}{5}.$$

Answer 3

Broad Steps

(1) We have the dr's of the two given lines

(2) Let P & Q be points on the two lines so that PQ is the shortest distance. Find equation of L through PQ

(3) Find distance of L from origin.

Details

Note that $\color{blue}{(2,3,-10)}$ and $\color{blue}{(4,-3,-2)}$ are direction ratios of the 2 given lines.

Let $P(2r_1-2, 3r_1 -6, -10r_1 +34)$ and $Q(4r_2-6, -3r_2 + 7, -2r_2 +7)$ be the points on the given lines so that $PQ$ is the line of shortest distance between the two lines.

The direction ratio's of $PQ$ are $(2r_1 - 4r_2 +4, 3r_1 + 3r_2 - 13, -10r_1 + 2r_2 +27)$

Since it is perpendicular to the 2 given lines, we will have

$$\color{blue}{2}(2r_1 - 4r_2 +4) + \color{blue}{3}(3r_1 + 3r_2 - 13) \color{blue}{-10}(-10r_1 + 2r_2 +27) = 0$$ and $$\color{blue}{4}(2r_1 - 4r_2 +4) \color{blue}{-3(}3r_1 + 3r_2 - 13) \color{blue}{-2}(-10r_1 + 2r_2 +27) = 0$$

These simplify to $113r_1 - 19r_2 - 301 = 0$ and $19r_1 - 29r_2 + 1 = 0$ which gives $\color{blue}{r_1 = 3, r_2 = 2}$

So we now have $P(4,3,4)$ and $Q(2,1,3)$ and direction ratios of $PQ$ as $2,2,1$

Finally, the equation of the line through $PQ$, or line $L$ is

$$\color{blue}{\frac{x-4}{2} = \frac{y-3}{2} = \frac{z-4}{1}}$$

The square of the distance of this line from the origin can be deduced using standard formulae.

However, in order to complete the solution, we can do the following: $(2r +4, 2r + 3, r +4)$ is any point on this line $L$. The square of its distance from the origin is clearly $(2r+4)^2 + (2r+3)^2 + (r+4)^2 = 9(r+2)^2 + 5$ which is minimum when $r = -2$ and the square of the distance is then $\color{blue}{5}$