The shortest distance between the lines x-7/3 = y+4/-16 = z-6/7 and x-10/3 = y-30/8 = 4-z/5 is

Correct Answer is : 28821288/2121288 units

Correct Answer is : 28821288/2121288 units

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VITEEE 2015 Paper

Section: Mathematics

Question : 114 of 120

Marks: +1, -0

\frac{x-7}{3}

\frac{y+4}{-16}

\frac{z-6}{7}

\frac{x-10}{3}

\frac{y-30}{8}

\frac{4-z}{5}

Solution:

Given, lines are

\frac{x-7}{3}

\frac{y+4}{-16}

\frac{z-6}{7}

and

\frac{x-10}{3}

\frac{y-30}{8}

\frac{4-z}{5}

The vector form of given lines are

r =

7\hat{i} - 4\hat{j} + 6\hat{k}

+ λ

(3\hat{i} - 16\hat{j} + 7\hat{k})

and r =

10\hat{i} + 30\hat{j} + 4\hat{k}

+ µ

(3\hat{i} + 8\hat{j} - 5\hat{k})

On comparing these equations with

r =

a_1 + \lambda b_1

and r =

a_2 + \mu b_2

we get

\vec{a_1}

7\hat{i} - 4\hat{j} + 6\hat{k}

\vec{a_2}

10\hat{i} + 30\hat{j} + 4\hat{k}

\vec{b_1}

3\hat{i} - 16\hat{j} + 7\hat{k}

and

\vec{b_2}

3\hat{i} + 8\hat{j} - 5\hat{k}

Shortest distance =

\left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|b_1 \times b_2|} \right|

\left| \frac{72+1224-144}{84} \right|

\left| \frac{1152}{84} \right|

\frac{288}{21}

units

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