Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mu=$

share|cite|improve this question
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

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 $\mu$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.