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I have 2 Question on $3-D$ Geometry

(1) The point on the Line $\displaystyle \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ which is Nearest to the Line

$\displaystyle \frac{x+1}{7}=\frac{y+1}{6}=\frac{z+1}{1}$ is

(2) If a Plane Contain $3-$ Lines drawn through $(1,1,1)$ and has a direction Ratios

$(1,-4,-1)\;\;,(3,5,7)$ and $(2,9,\mu)$. Then value of $\mu=$

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For the first, you could try parametrizing both lines and minimizing distance turns into a calculus problem. For the second, $(2,9,\mu)$ must be linearly dependent with $(1,-4,1)$ and $(3,5,7)$ or you get a three-dim space instead of two. This should allow you to calculate $\mu$. –  user45150 Jan 25 '13 at 18:27
    
$\displaystyle \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}=\lambda$ we get $x=\lambda+3\;,y=-2\lambda+5\;,z=\lambda+7$ Similarly for other $\displaystyle \frac{x+1}{7}=\frac{y+1}{6}=\frac{z+1}{1}=\mu$, we get $x=7\mu-1\;,y=6\mu-1\;,z=\mu-1$ Now $\displaystyle K=d^2_{Min.}=$ –  juantheron Jan 25 '13 at 18:30
    
Right, then try and minimize the distance between the two (or equivalently minimize the distance squared). ie minimize $((\lambda+3)-(7\mu-1))^2+((-2\lambda+5)-(6\mu-1))^2+((\lambda+7)-(\mu-1))^2$ –  user45150 Jan 25 '13 at 18:35
    
Now $\displaystyle K=d^2_{Min.}=(7\mu-\lambda-4)^2+(6\mu+2\lambda-6)^2+(\mu-\lambda-8)^2$ now after that How can i minimize the function which Involve two variable i.e $\lambda$ and $\mu$ . Thanks –  juantheron Jan 25 '13 at 18:38
    
The minimum will have to occur when $\frac{\partial K}{\partial \mu}$ and $\frac{\partial K}{\partial \lambda}$ are both zero, so take these derivatives and see when the are both zero, and check if it is a minimum there. –  user45150 Jan 25 '13 at 18:40
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1 Answer

For part 1, see this page.

For the second part: The equation of the three lines are $$\frac{x-1}{l_i}+\frac{y-1}{m_i}+ \frac{z-1}{n_i}, i = 1,2,3 $$ where $(l_i,m_i,n_i)$ is the direction ratio for the line.

Let the equation of the plane containing these three lines be $ax+by+cz =1 $. The plane contains the point $(1,1,1)$. So, $$a+b+c =1 $$ Also, $(a,b,c)$ is the direction ratio of the normal to the plane. For any line in the plane with direction ratio $(l,m,n)$$$al+bm+cn = 0$$

So, $$a.1+b.(-4)+c.(-1) =0$$ $$a.3+b.5+c.7 =0$$ Use these three equations to find out $a,b,c$. Then, use $a.2+b.9+c.\mu =0$ to find out $\mu$.

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Thanks dextro for (I) Using Skew lines Method ............................... line vector which is perpendicular to these line is $\vec{n} = n_{1} \times n_{2}$ where $\vec{n_{1}}=\vec{i}-2\vec{j}+\vec{k}$ and $\vec{n_{1}}=7\vec{i}-6\vec{j}+\vec{k}$ so we get $\vec{n} = 4\vec{i}+6\vec{j}+8\vec{k}$ now after that how can i proceed further thanks –  juantheron Jan 25 '13 at 19:24
    
Find out any point on line one (a), any point on line 2 (c) and apply $d = |n.(c-a)|$ –  dexter04 Jan 25 '13 at 20:45
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