Name: Angle Between Two Lines (Vector Form)
Uploaded: 2019-07-19
Duration: 5 min 27 s
Description: How to find the angle between two lines in coordinate geometry using slope? Here we use the direction equivalent of the same to find the angle between two li...

Question

Find the length and equation of the line of the shortest distance between the lines

3 x - 9 y + 5 z = 0 = x + y - z

and

6 x + 8 y + 3 z - 13 = 0 = x + 2 y + z - 3

.

Accepted Answer

We have to find the length and equation of the line of the shortest distance between the lines $3 x - 9 y + 5 z = 0 = x + y - z$ and $6 x + 8 y + 3 z - 13 = 0 = x + 2 y + z - 3$ .

We have to convert these equations into symmetric form

$3 x - 9 y + 5 z = 0$

$x + y - z = 0$

$\frac{x}{9 - 5} = \frac{- y}{- 3 - 5} = \frac{z}{3 + 9}$

$\Rightarrow \frac{x}{1} = \frac{y}{2} = \frac{z}{3} = a . . . . . (1)$

Now, $3 x - 9 y + 5 z = 0 = x + y - z$

if $b$ is a vector along the line then $b = n_{1} \times n_{2}$ ,

$b = |\begin{matrix} i & j & k \\ 6 & 8 & 3 \\ 1 & 2 & 1 \end{matrix}| = (8 - 6) i - (6 - 3) j + (12 - 8) k$

$b = 2 i - 3 j + 4 k$

Let $z = 0$ , then line passes through any point $P (p, q, 0)$ satisfy the equations $(1)$ and $(2)$

$\Rightarrow 6 p + 8 q - 13 = 0 . . . . . (2), p + 2 q - 3 = 0 . . . . . (3)$

By solving equation $(3)$ and $(4)$

$q = \frac{5}{4}$

Putting $q$ , value in equation $(3)$ , we get $p = \frac{1}{2}$

Therefore, symmetric form of line is $\Rightarrow \frac{x - \frac{1}{2}}{2} = \frac{y - \frac{5}{4}}{- 3} = \frac{z}{4} = b . . . . . (4)$ .

Any point on line $(1)$ is $P (a, 2 a, 3 a)$ , and on line $(2)$ is $Q (2 b + \frac{1}{2}, - 3 b + \frac{5}{4}, 4 b)$ .

If $P Q$ is the line of the shortest distance, then direction ratios of $P Q = (2 b + \frac{1}{2} - a, - 3 b + \frac{5}{4} - 2 a, 4 b - 3 a)$ .

As $P Q$ is perpendicular to lines $(1)$ and $(4)$ ,

Therefore $1 (2 b + \frac{1}{2} - a) + 2 (- 3 b + \frac{5}{4} - 2 a) + 3 (4 b - 3 a) = 0$

$\Rightarrow 8 b - 14 a + 3 = 0 . . . . . (5)$

And $2 (2 b + \frac{1}{2} - a) - 3 (- 3 b + \frac{5}{4} - 2 a) + 4 (4 b - 3 a) = 0$

$\Rightarrow 29 b + - 4 a - \frac{11}{4} = 0 . . . . . (6)$

On solving equations $(3)$ and $(4)$ , we get $b = \frac{- 53}{348}$ , $a = \frac{7}{174}$ .

So, points $P (\frac{7}{174}, \frac{14}{174}, \frac{21}{174})$ and $Q (\frac{121}{348}, \frac{276}{348}, \frac{- 212}{348})$ .

Therefore, length of the shortest distance $P Q = \sqrt{{(\frac{121}{348} - \frac{7}{174})}^{2} + {(\frac{276}{348} - \frac{14}{174})}^{2} + {(- \frac{212}{348} - \frac{21}{174})}^{2}}$

$P Q = \frac{11}{\sqrt{342}}$

Direction ratios of shortest distance line are $283, - 342$ and $443$ .

Therefore, equation of the shortest distance line is $\frac{x - \frac{3161}{1197}}{283} = \frac{y - \frac{1090}{1197}}{- 342} = \frac{z - 0}{443}$

$\Rightarrow 10 x - 29 y + 16 z = 013 x + 82 y + 55 z - 109$ .

Find the length and equation of the line of the shortest distance between the lines $3 x - 9 y + 5 z = 0 = x + y - z$ and $6 x + 8 y + 3 z - 13 = 0 = x + 2 y + z - 3$ .

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the length and equation of the line of the shortest distance between the lines 3x-9y+5z=0=x+y-z and 6x+8y+3z-13=0=x+2y+z-3.

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the length and equation of the line of the shortest distance between the lines $3 x - 9 y + 5 z = 0 = x + y - z$ and $6 x + 8 y + 3 z - 13 = 0 = x + 2 y + z - 3$ .