Mathematics : Three Dimensional Geometry

Show that lines:

Also, find the equation of the plane containing these lines. 

Find the value of p, so that the lines:
are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, -4) and parallel to line l₁.

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x- y + z = 0. Also find the distance of the plane obtained above, from the origin.

A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of 80 on each piece of type A and 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

Find the Cartesian equation of the line passes through the point (-2, 4, -5) and is parallel to the line 

Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.

Find the shortest distance between the lines.

$\overrightarrow{\mathrm{r}} = (4\hat{\mathrm{i}} -\hat{\mathrm{j}}) + \mathrm{\lambda } (\hat{\mathrm{i}} - \hat{\mathrm{j}} + 2\hat{\mathrm{k}}) + \mathrm{\mu } (2\hat{\mathrm{i}} + 4\hat{\mathrm{j}}-5\widehat{\mathrm{k})}$

Find the distance of the point (-1,-51-10) from the point of intersection of the line $\overrightarrow{\mathrm{r}} = 2\hat{\mathrm{i}} -\hat{\mathrm{j}} + 2\hat{\mathrm{k}} + \mathrm{\lambda } (3\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + 2\hat{\mathrm{k}})$ and the plane $\overrightarrow{r}. (\hat{i}-\hat{j} + \hat{k}) = 5$

Find the shortest distance between the following lines:

$\frac{\mathrm{x} - 3}{1} = \frac{\mathrm{y} - 5}{-2} = \frac{\mathrm{z} - 7}{1} \mathrm{and} \frac{{\displaystyle \mathrm{x} + 1}}{{\displaystyle 7}} = \frac{{\displaystyle \mathrm{y} + 1}}{{\displaystyle -6}} = \frac{{\displaystyle \mathrm{z} + 1}}{{\displaystyle 1}}$

Find the point on the line $\frac{\mathrm{x} + 2}{3} = \frac{\mathrm{y} + 1}{2} = \frac{\mathrm{z} - 3}{2}$ at a distance 3 $\sqrt{2}$ from the point
(1, 2, 3).