If is the shortest distance between the lines x + 1 = 2y = –12z, x = y + 2 = 6z – 6 and is the shortest distance between the lines , then the value of is ___________. [2024]

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Solution

Q.
If $d_{1}$ is the shortest distance between the lines x + 1 = 2y = –12z, x = y + 2 = 6z – 6 and $d_{2}$ is the shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}, \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ , then the value of $\frac{32 \sqrt{3} d_{1}}{d_{2}}$ is ___________. [2024]

Ans.
(16)

We have, $l_{1}$ : x + 1 = 2y = –12z and $l_{2}$ : x = y + 2 = 6z – 6

So, $l_{1}$ : $\frac{x - (- 1)}{1} = \frac{y - 0}{\frac{1}{2}} = \frac{z - 0}{\frac{- 1}{12}}$ and $l_{2}$ : $\frac{x - 0}{1} = \frac{y - (- 2)}{1} = \frac{z - 1}{\frac{1}{6}}$

These lines can be written in vector form as

${\vec{r}}_{1} = (- \hat{i} + 0 \hat{j} + 0 \hat{k}) + λ (\hat{i} + \frac{\hat{j}}{2} - \frac{\hat{k}}{12})$ and ${\vec{r}}_{2} = (0 \hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} + \hat{j} + \frac{1}{6} \hat{k})$

For, $\vec{r_{1}} = \vec{a_{1}} + λ \vec{b_{1}}$ and $\vec{r_{2}} = \vec{a_{2}} + λ \vec{b_{2}}$

Shortest distance between $l_{1}$ and $l_{2}$ is given by

$d_{1} =$ $| \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$ $=$ $| \frac{(\hat{i} + 2 \hat{j} + \hat{k}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 / 2 & - 1 / 12 \\ 1 & 1 & 1 / 6 \end{matrix} |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$

$=$ $| \frac{(\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\frac{1}{6} \hat{i} + \frac{1}{4} \hat{j} + \frac{1}{2} \hat{k})}{\sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{4}}} |$ $=$ $| \frac{\frac{1}{6} + \frac{1}{2} + \frac{1}{2}}{\sqrt{\frac{49}{144}}} |$ $= \frac{7}{6} \times \frac{12}{7} = 2$

Similarly, $l_{3}$ : $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $l_{4}$ : $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$

These lines can be written in vector form as

${\vec{r}}_{1} = (\hat{i} - 8 \hat{j} + 4 \hat{k}) + λ (2 \hat{i} - 7 \hat{j} + 5 \hat{k})$

${\vec{r}}_{2} = (\hat{i} + 2 \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} - 3 \hat{k})$

Shortest distance between $l_{3}$ and $l_{4}$ is given as

$d_{2} =$ $| \frac{(10 \hat{j} + 2 \hat{k}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} |}{| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} | |} |$

$=$ $| \frac{(10 \hat{j} + 2 \hat{k}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k})}{\sqrt{16^{2} + 16^{2} + 16^{2}}} |$ $= \frac{160 + 32}{16 \sqrt{3}} = \frac{12}{\sqrt{3}} = 4 \sqrt{3}$

Now, $\frac{32 \sqrt{3} d_{1}}{d_{2}} = \frac{32 \sqrt{3} \times 2}{4 \sqrt{3}} = 16$ .

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