Find the shortest distance between the following pairs of lines whose vector equations are-i r -3 i -8 j -3 k - 3 i - j - k - and r -3 i -7 j -6 k -3 i -2 j -4 k -ii r -3 i -5 j -7 k - i -2 j -7 k - and r - i - j - k - 7 i -6 j - k -iii r - i -2 j -3 k - 2 i -3 j -4 k - and r -2 i -4 j -5 k - 3 i -4 j-5 k -iv r -1- t i - t -2 j -3- t k - and r - s -1 i -2 s -1 j -2 s -1 k -v r -1 i -1 j -1- k - and r -1- i -2 -1 j -2 k -vi r -2 i - j - k - i -5 j -2 k - and- r - i -2 j - k - i - j - k -vii r - i - j - 2 i - j - k - and- r -2 i - j - k - 3 i -5 j -2 k -

(i) r #8594;=3i^+8j^+3k^+ #955;3i^ j^+k^ #160;and #160;r #8594;= 3i^ 7j^+6k^+ #956; 3i^+2j^+4k^ Comparing the given equations with the equations r ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Vector Triple Product

Find the shor...

Question

Find the shortest distance between the following pairs of lines whose vector equations are:

(i) $\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})$

(ii) $\vec{r} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 7 \hat{k}) and \vec{r} = - \hat{i} - \hat{j} - \hat{k} + μ (7 \hat{i} - 6 \hat{j} + \hat{k})$

(iii) $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 5 \hat{k})$

(iv) $\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - t) \hat{k} and \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}$

(v) $\vec{r} = (λ - 1) \hat{i} + (λ + 1) \hat{j} - (1 + λ) \hat{k} and \vec{r} = (1 - μ) \hat{i} + (2 μ - 1) \hat{j} + (μ + 2) \hat{k}$

(vi) $\vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 2 \hat{k}) and, \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$

(vii) $\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} + \hat{j} - \hat{k} + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$

(viii) $\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}$ and $\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})$ [NCERT EXEMPLAR]

Open in App

Solution

(i) $\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \vec{a_{2}} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \vec{b_{1}} = 3 \hat{i} - \hat{j} + \hat{k} \vec{b_{2}} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix}| = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 6)}^{2} + {(- 15)}^{2} + 3^{2}} = \sqrt{36 + 225 + 9} = \sqrt{270} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (- 6 \hat{i} - 15 \hat{j} + 3 \hat{k}) . (- 6 \hat{i} - 15 \hat{j} + 3 \hat{k}) = 36 + 225 + 9 = 270$

The shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = \frac{270}{\sqrt{270}} = \sqrt{270}$

(ii) $\vec{r} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 7 \hat{k}) and \vec{r} = - \hat{i} - \hat{j} - \hat{k} + μ (7 \hat{i} - 6 \hat{j} + \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \vec{a_{2}} = - \hat{i} - \hat{j} - \hat{k} \vec{b_{1}} = \hat{i} - 2 \hat{j} + 7 \hat{k} \vec{b_{2}} = 7 \hat{i} - 6 \hat{j} + \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 7 \\ 7 & - 6 & 1 \end{matrix}| = 40 \hat{i} + 48 \hat{j} + 8 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{40^{2} + 48^{2} + 8^{2}} = \sqrt{1600 + 2304 + 64} = \sqrt{3968} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (- 4 \hat{i} - 6 \hat{j} - 8 \hat{k}) . (40 \hat{i} + 48 \hat{j} + 8 \hat{k}) = - 160 - 288 - 64 = - 512$

The shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{- 512}{\sqrt{3968}}| = \frac{512}{\sqrt{3968}}$

(iii) $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 5 \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{a_{2}} = 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \vec{b_{1}} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \vec{b_{2}} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix}| = - \hat{i} + 2 \hat{j} - \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 1)}^{2} + 2^{2} + {(- 1)}^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (\hat{i} + 2 \hat{j} + 2 \hat{k}) . (- \hat{i} + 2 \hat{j} - \hat{k}) = - 1 + 4 - 2 = 1$

The shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{1}{\sqrt{6}}| = \frac{1}{\sqrt{6}}$

(iv) $\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - t) \hat{k} and \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}$

The vector equations of the given lines can be re-written as

$\vec{r} = \hat{i} - 2 \hat{j} + 3 \hat{k} + t (- \hat{i} + \hat{j} - \hat{k}) and \vec{r} = \hat{i} - \hat{j} - \hat{k} + s (\hat{i} + 2 \hat{j} - 2 \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = \hat{i} - 2 \hat{j} + 3 \hat{k} \vec{a_{2}} = \hat{i} - \hat{j} - \hat{k} \vec{b_{1}} = - \hat{i} + \hat{j} - \hat{k} \vec{b_{2}} = \hat{i} + 2 \hat{j} - 2 \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = \hat{j} - 4 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & - 1 \\ 1 & 2 & - 2 \end{matrix}| = - 3 \hat{j} - 3 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (\hat{j} - 4 \hat{k}) . (- 3 \hat{j} - 3 \hat{k}) = - 3 + 12 = 9$

The shortest distance between the line $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by
$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{9}{3 \sqrt{2}}| = \frac{3}{\sqrt{2}}$

(v) $\vec{r} = (λ - 1) \hat{i} + (λ + 1) \hat{j} - (1 + λ) \hat{k} and \vec{r} = (1 - μ) \hat{i} + (2 μ - 1) \hat{j} + (μ + 2) \hat{k}$

The vector equations of the given lines can be re-written as

$\vec{r} = - \hat{i} + \hat{j} - \hat{k} + λ (\hat{i} + \hat{j} - \hat{k}) and \vec{r} = \hat{i} - \hat{j} + 2 \hat{k} + μ (- \hat{i} + 2 \hat{j} + \hat{k})$
Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = - \hat{i} + \hat{j} - \hat{k} \vec{a_{2}} = \hat{i} - \hat{j} + 2 \hat{k} \vec{b_{1}} = \hat{i} + \hat{j} - \hat{k} \vec{b_{2}} = - \hat{i} + 2 \hat{j} + \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = 2 \hat{i} - 2 \hat{j} + 3 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & - 1 \\ - 1 & 2 & 1 \end{matrix}| = 3 \hat{i} + 3 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{3^{2} + 3^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (2 \hat{i} - 2 \hat{j} + 3 \hat{k}) . (3 \hat{i} + 3 \hat{k}) = 6 + 9 = 15$

The shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{15}{3 \sqrt{2}}| = \frac{5}{\sqrt{2}}$

(vi) $\vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 2 \hat{k}) and, \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = 2 \hat{i} - \hat{j} - \hat{k} \vec{a_{2}} = \hat{i} + 2 \hat{j} + \hat{k} \vec{b_{1}} = 2 \hat{i} - 5 \hat{j} + 2 \hat{k} \vec{b_{2}} = \hat{i} - \hat{j} + \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = - \hat{i} + 3 \hat{j} + 2 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 5 & 2 \\ 1 & - 1 & 1 \end{matrix}| = - 3 \hat{i} + 3 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 3)}^{2} + 3^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (- \hat{i} + 3 \hat{j} + 2 \hat{k}) . (- 3 \hat{i} + 3 \hat{k}) = 3 + 6 = 9$

The shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{9}{3 \sqrt{2}}| = \frac{3}{\sqrt{2}}$

(vii) $\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} + \hat{j} - \hat{k} + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = \hat{i} + \hat{j} \vec{a_{2}} = 2 \hat{i} + \hat{j} - \hat{k} \vec{b_{1}} = 2 \hat{i} - \hat{j} + \hat{k} \vec{b_{2}} = 3 \hat{i} - 5 \hat{j} + 2 \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = \hat{i} - \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 1 & 1 \\ 3 & - 5 & 2 \end{matrix}| = 3 \hat{i} - \hat{j} - 7 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{3^{2} + {(- 1)}^{2} + {(- 7)}^{2}} = \sqrt{9 + 1 + 49} = \sqrt{59} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (\hat{i} - \hat{k}) . (3 \hat{i} - \hat{j} - 7 \hat{k}) = 3 + 7 = 10$

The shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{10}{\sqrt{59}}| = \frac{10}{\sqrt{59}}$

(viii) The vector equations of the given lines can be re-written as

$\vec{r} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} + λ (3 \hat{i} - 16 \hat{j} + 7 \hat{k})$ and $\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})$

Comparing the given equations with the equations $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ , we get

$\vec{a_{1}} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k}$

$\vec{b_{1}} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k}$

$\vec{a_{2}} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k}$

$\vec{b_{2}} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k}$

$∴ \vec{a_{2}} - \vec{a_{1}} = (15 \hat{i} + 29 \hat{j} + 5 \hat{k}) - (8 \hat{i} - 9 \hat{j} + 10 \hat{k}) = 7 \hat{i} + 38 \hat{j} - 5 \hat{k}$

$\vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & 8 & - 5 \end{matrix}| = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{24^{2} + 36^{2} + 72^{2}} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84$

Also,

$(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (7 \hat{i} + 38 \hat{j} - 5 \hat{k}) . (24 \hat{i} + 36 \hat{j} + 72 \hat{k}) = 7 \times 24 + 38 \times 36 + (- 5) \times 72 = 168 + 1368 - 360 = 1176$

We know that the shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by $d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}|$ .

∴ Required shortest distance between the given pairs of lines,

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{1176}{84}| = 14$

Suggest Corrections