Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

Consider the lines

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y + 2} \over 2} = {{z - 3} \over 3}$$

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The distance of the point $$(1, 1, 1)$$ from the plane passing through the point $$(-1, -2, -1)$$ and whose normal is perpendicular to both the lines $${L_1}$$ and $${L_2}$$ is :

Let two non-collinear unit vectors $$\widehat a$$ and $$\widehat b$$ form an acute angle. A point $$P$$ moves so that at any time $$t$$ the position vector $$\overrightarrow {OP} $$ (where $$O$$ is the origin) is given by $$\widehat a\cos t + \widehat b\sin t.$$ When $$P$$ is farthest from origin $$O,$$ let $$M$$ be the length of $$\overrightarrow {OP} $$ and $$\widehat u$$ be the unit vector along $$\overrightarrow {OP} $$. Then :