Find the shortest distance between the linesi r - i -2 j - k - i - j - k - and- r -2 i - j - k - 2 i - j -2 k -ii x-17-y-1-6-z-11 and x-31-y-5-2-z-71iii r - i -2 j -3 k - i -3 j -2 k - and r -4 i -5 j -6 k - 2 i -3 j - k -iv r -6 i -2 j -2 k - i -2 j -2 k - and r -4 i - k - 3 i -2 j -2 k -

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

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

Find the shor...

Question

Find the shortest distance between the lines
(i) $\vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$

(ii) $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$

(iii) $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$

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

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Solution

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

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

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

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

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

(ii) $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$

Since the first line passes through the point ( $-$ 1, $-$ 1, $-$ 1) and has direction ratios proportional to 7, $-$ 6, 1, its vector equation is
$\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} Here, \vec{a_{1}} = - \hat{i} - \hat{j} - \hat{k} \vec{b_{1}} = 7 \hat{i} - 6 \hat{j} + \hat{k}$

Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1, $-$ 2, 1.
Its vector equation is
$\vec{r} = \vec{a_{2}} + μ \vec{b_{2}} Here, \vec{a_{2}} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \vec{b_{2}} = \hat{i} - 2 \hat{j} + \hat{k}$

Now,
$\vec{a_{2}} - \vec{a_{1}} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & - 6 & 1 \\ 1 & - 2 & 1 \end{matrix}| = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 4)}^{2} + {(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{16 + 36 + 64} = \sqrt{116} and (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (4 \hat{i} + 6 \hat{j} + 8 \hat{k}) . (- 4 \hat{i} - 6 \hat{j} - 8 \hat{k}) = - 16 - 36 - 64 = - 116$

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

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| \Rightarrow d = |\frac{- 116}{\sqrt{116}}| = \sqrt{116} = 2 \sqrt{29}$

(iii) $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$

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
$\vec{a_{1}} = \hat{i} + 2 \hat{j} + 3 \hat{k} \vec{a_{2}} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} \vec{b_{1}} = \hat{i} - 3 \hat{j} + 2 \hat{k} \vec{b_{2}} = 2 \hat{i} + 3 \hat{j} + \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 2 \\ 2 & 3 & 1 \end{matrix}| = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 9)}^{2} + 3^{2} + 9^{2}} = \sqrt{81 + 9 + 81} = \sqrt{171} and (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) . (- 9 \hat{i} + 3 \hat{j} + 9 \hat{k}) = - 27 + 9 + 27 = 9$

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
$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| \Rightarrow d = |\frac{9}{\sqrt{171}}| = \frac{3}{\sqrt{19}}$

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

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
$\vec{a_{1}} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} \vec{a_{2}} = - 4 \hat{i} - \hat{k} \vec{b_{1}} = \hat{i} - 2 \hat{j} + 2 \hat{k} \vec{b_{2}} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = - 10 \hat{i} - 2 \hat{j} - 3 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2 \end{matrix}| = 8 \hat{i} + 8 \hat{j} + 4 \hat{k} \Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{8^{2} + 8^{2} + 4^{2}} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 and (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (- 10 \hat{i} - 2 \hat{j} - 3 \hat{k}) . (8 \hat{i} + 8 \hat{j} + 4 \hat{k}) = - 80 - 16 - 12 = - 108$

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

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| \Rightarrow d = |\frac{- 108}{12}| = 9$

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Similar questions

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(i)

\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - 2 \hat{k}) and \vec{r} = \hat{i} - \hat{j} + 2 \hat{k} - μ (2 \hat{i} + 4 \hat{j} - 4 \hat{k})

(ii)

\vec{r} = (3 \hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 2 \hat{k}) and \vec{r} = (5 \hat{j} - 2 \hat{k}) + μ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})

(iii)

\vec{r} = λ (\hat{i} + \hat{j} + 2 \hat{k}) and \vec{r} = 2 \hat{j} + μ \{(\sqrt{3} - 1) \hat{i} - (\sqrt{3} + 1) \hat{j} + 4 \hat{k}\}

Q. Find the shortest distance between the skew lines

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and

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

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\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})

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(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]

Q. The angle between the lines

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Q. Find the shortest distance between the following pairs of parallel lines whose equations are:
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\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) and \vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + μ (- \hat{i} + \hat{j} - \hat{k})

(ii)

\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 (4 \hat{i} - 2 \hat{j} + 2 \hat{k})

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Find the shortest distance between the lines (i) r→=i^+2j^+k^+λi^-j^+k^ and, r→=2i^-j^-k^+μ2i^+j^+2k^ (ii) x+17=y+1-6=z+11 and x-31=y-5-2=z-71 (iii) r→=i^+2j^+3k^+λi^-3j^+2k^ and r→=4i^+5j^+6k^+μ2i^+3j^+k^ (iv) r→=6i^+2j^+2k^+λi^-2j^+2k^ and r→=-4i^-k^+μ3i^-2j^-2k^