Question

The shortest distance between the two lines

Answer 1

Answer:

GIVEN LINES :–

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$

AND —

$\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{ z - 5 }{5}$

WRITE THESE LINES IN VECTOR FORM —

FORM OF LINE —

$l = a + \alpha ( b)$

LINES :–

$l_{1} = (i + 2j + 3k) + \alpha (2i + 3j + 4k)$

$l_{2} = (2i + 4j + 5k) + \alpha (3i + 4j + 5k)$

SHORTEST DISTANCE :–

$d = \frac{( a_{2} - a_{1}).( b_{1} \times b_{2})}{ | b_{1} \times b_{2}| }$

$d = \frac{(i + 2j + 2k).( - i + 2j - k)}{ |( - i + 2j - k) | }$

$d = \frac{ - 1 + 4 - 2}{ \sqrt{1 + 4 + 1} }$

$d = \frac{1}{ \sqrt{6} } unit$

HOPE YOU LIKE IT....

Answer 2

Answer:

GIVEN LINES :–

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} 2 x−1 = 3

y−2 = 4

z−3

AND —

\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{ z - 5 }{5} 3x−2 = 4

y−4 = 5

z−5

WRITE THESE LINES IN VECTOR FORM —

FORM OF LINE —

l = a + \alpha ( b)l=a+α(b)

LINES :–

l_{1} = (i + 2j + 3k) + \alpha (2i + 3j + 4k)l

1

=(i+2j+3k)+α(2i+3j+4k)

l_{2} = (2i + 4j + 5k) + \alpha (3i + 4j + 5k)l

2

=(2i+4j+5k)+α(3i+4j+5k)

SHORTEST DISTANCE :–

d = \frac{( a_{2} - a_{1}).( b_{1} \times b_{2})}{ | b_{1} \times b_{2}| }d=

∣b

1 ×b 2 ∣(a 2 −a 1 ).(b 1 ×b )

d = \frac{(i + 2j + 2k).( - i + 2j - k)}{ |( - i + 2j - k) | }d= ∣(−i+2j−k)∣

(i+2j+2k).(−i+2j−k)

d = \frac{ - 1 + 4 - 2}{ \sqrt{1 + 4 + 1} }d= 1+4+1−1+4−2

d = \frac{1}{ \sqrt{6} } unitd= 61unit