Question

Find the distance between parallel lines. x/2 = y/- 1 = z/2 and and (x - 1)/2 = (y - 1)/- 1 = (z - 3)/2 .​

Answer 1

Step-by-step-Explanation

Given:

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

To find: Find the distance between given parallel lines.

Solution:

Shortest distance between two parallel lines is given by

$\bold{\red{D = \frac{ |(a_2 - a_1) \times b| }{ |b| } }}$

here this is in vector form.

a is position vector and b is direction vector.

Step 1: Write vector form of equations

Because line 1 is passing through origin,so

$r_1= 0 + \lambda(2 \hat i - \hat j + 2\hat k) \\ \\ r_2= ( \hat i + \hat j + 3\hat k) + \lambda(2 \hat i - \hat j + 2\hat k)$

here

$a_1 = 0 \hat i + 0 \hat j + 0\hat k \\ \\ a_2 = \hat i + \hat j + 3\hat k \\ \\ b = 2\hat i - \hat j + 2\hat k$

Step 2: Put these values in formula

$D= \frac{ |( \hat i + \hat j + 3\hat k) \times (2\hat i - \hat j + 2\hat k)| }{ \sqrt{4 + 1 + 4} }$

Step 3: Take cross product of vectors in numerator

$( \hat i + \hat j + 3\hat k) \times (2\hat i - \hat j + 2\hat k)\\\\=\left|\begin{array}{ccc}\hat i&\hat j&\hat j\\1&1&3\\2&-1&2\\\end{array}\right|$

$| \hat i(2 + 3) - \hat j(2 - 6) + \hat k( - 1 - 2)| \\ \\ = | 5\hat i + 4 \hat j - 3 \hat k| \\ \\ = \sqrt{25 + 16 + 9} \\\\ = \sqrt{50}$

Thus,

Distance

$D = \frac{ \sqrt{50} }{3} \\ \\ D = \frac{5 \sqrt{2} }{3} \: units$

Final answer:

Distance between given parallel lines is $\bold{\red{\frac{5 \sqrt{2} }{3}}} \: units$

Hope it helps you.

To learn more:

1) Prove that the vectors (2, 1, 4), (1, –1, 2) and (3, 1, –2) for a basis of V3

(R).

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2) using laplace transform solve y'+5y'+6y=2 given y(0)=0 y'(0)=0

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