Lecture 5 Three Dimensional Geometry

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Lecture - 5 Chapter 11 Three Dimensional Geometry

“Whenever you find yourself on the side of the majority, it is time to pause and reflect.” –Mark Twain

In this lecture I am discussing skew lines in the three dimensional geometry.

1. Find the shortest distance between the lines & check whether lines are intersecting or not:

b. \(\frac{{x+1}}{7}=\frac{{y+1}}{{-6}}=\frac{{z+1}}{1}\) & \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
c. \(r={i+2j+3k}+\lambda(i-3j+2k)\) & \( r=4i+5j+6k+\mu(2i+3j+k)\)
d. \(r=(1-t)i+(t-2)j+(3-2t)k\) & \(r=(s+1)i+(2s-1)j-(2s+1)k \)
e. \(r=(6i+2j+2k)+\lambda(i-2j+2k)\) & \(r=-4i-k+\mu(3i-2j-2k)\)

2. Find the shortest distance between the lines and hence write whether the lines are intersecting or not : \(\frac{x-1}{2}=\frac{y+1}{3}=z\) & \(\frac{x+1}{5}=\frac{y-2}{1}=\frac{z}{2}\) . 

3. Show that the lines \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\) & \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\) are coplanar.

4. Show that the lines \(\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}\) & \(\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}\) are coplanar.

5. Find the distance between lines
\(r=i+2j+k+\lambda(2i+3j+6k)\) & \(r=3i+3j-5k+\mu(2i+3j+6k)\).

6. Find the distance between following pairs of lines:
\(r={i+2j+3k}+\lambda(i-j+k)\) & \(r={2i-j-k}+\mu(-i+j-k)\)
\(r={i+j}+\lambda(2i-j+k)\) & \(r={2i+j-k}+\mu(4i-2j+2k)\)

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