By computing the shortest distance determine whether the following pairs of lines intersect or not-i r - i - j - 2 i - k - and r -2 i - j - i - j - k -ii r - i - j - k - 3 i - j - and r -4 i - k - 2 i -3 k -iii x-12-y-13-z and x-15-y-21 - z-2iv x-54-y-7-5-z-3-5 and x-87-y-71-z-53

(i) r #8594;=i^ j^+ #955;2i^+k^ #160;and #160;r #8594;=2i^ j^+ #956;i^+j^ k^ Comparing the given equations with the equations r #8594;=a1 #8594 ...

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Byju's Answer

Standard XII

Mathematics

Solving System of Linear Equations Using Inverse

By computing ...

Question

By computing the shortest distance determine whether the following pairs of lines intersect or not:
(i) $\vec{r} = (\hat{i} - \hat{j}) + λ (2 \hat{i} + \hat{k}) and \vec{r} = (2 \hat{i} - \hat{j}) + μ (\hat{i} + \hat{j} - \hat{k})$

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

(iii) $\frac{x - 1}{2} = \frac{y + 1}{3} = z and \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2$

(iv) $\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}$

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Solution

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

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

$\vec{a_{1}} = \hat{i} - \hat{j} \vec{a_{2}} = 2 \hat{i} - \hat{j} \vec{b_{1}} = 2 \hat{i} + \hat{k} \vec{b_{2}} = \hat{i} + \hat{j} - \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = \hat{i} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & - 1 \end{matrix}| = - \hat{i} + 3 \hat{j} + 2 \hat{k} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (\hat{i}) . (- \hat{i} + 3 \hat{j} + 2 \hat{k}) = - 1 We observe (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) \neq 0 Thus, the given lines do not intersect .$

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

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

$\vec{a_{1}} = \hat{i} + \hat{j} - \hat{k} \vec{a_{2}} = 4 \hat{i} - \hat{k} \vec{b_{1}} = 3 \hat{i} - \hat{j} \vec{b_{2}} = 2 \hat{i} + 3 \hat{k} ∴ \vec{a_{2}} - \vec{a_{1}} = 3 \hat{i} - \hat{j} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 0 \\ 2 & 0 & 3 \end{matrix}| = - 3 \hat{i} - 9 \hat{j} + 2 \hat{k} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (3 \hat{i} - \hat{j}) . (- 3 \hat{i} - 9 \hat{j} + 2 \hat{k}) = - 9 + 9 = 0 We observe (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = 0 Thus, the given lines intersect .$

(iii) $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 0}{1} and \frac{x + 1}{5} = \frac{y - 2}{1} = \frac{z - 2}{0}$

Since the first line passes through the point (1, $-$ 1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
$\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} . . . (1) Here, \vec{a_{1}} = \hat{i} - \hat{j} + 0 \hat{k} \vec{b_{1}} = 2 \hat{i} + 3 \hat{j} + \hat{k}$

Also, the second line passes through the point ( $-$ 1, 2, 2) and has direction ratios proportional to 5, 1, 0.
Its vector equation is
$\vec{r} = \vec{a_{2}} + μ \vec{b_{2}} . . . (2) Here, \vec{a_{2}} = - \hat{i} + 2 \hat{j} + 2 \hat{k} \vec{b_{2}} = 5 \hat{i} + \hat{j} + 0 \hat{k}$

Now,
$\vec{a_{2}} - \vec{a_{1}} = - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{matrix}| = - \hat{i} + 5 \hat{j} - 13 \hat{k} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (- 2 \hat{i} + 3 \hat{j} + 2 \hat{k}) . (- \hat{i} + 5 \hat{j} - 13 \hat{k}) = 2 + 15 - 26 = - 9 We observe (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) \neq 0 Thus, the given lines do not intersect .$

(iv) $\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}$

Since the first line passes through the point (5, 7, $-$ 3) and has direction ratios proportional to 4, $-$ 5, $-$ 5, its vector equation is
$\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} . . . (1) Here, \vec{a_{1}} = 5 \hat{i} + 7 \hat{j} - 3 \hat{k} \vec{b_{1}} = 4 \hat{i} - 5 \hat{j} - 5 \hat{k}$

Also, the second line passes through the point (8, 7, 5) and has direction ratios proportional to 7, 1, 3.
Its vector equation is
$\vec{r} = \vec{a_{2}} + μ \vec{b_{2}} . . . (2) Here, \vec{a_{2}} = 8 \hat{i} + 7 \hat{j} + 5 \hat{k} \vec{b_{2}} = 7 \hat{i} + \hat{j} + 3 \hat{k}$

Now,
$\vec{a_{2}} - \vec{a_{1}} = 3 \hat{i} + 8 \hat{k} and \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 5 & - 5 \\ 7 & 1 & 3 \end{matrix}| = - 10 \hat{i} - 47 \hat{j} + 39 \hat{k} (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (3 \hat{i} + 8 \hat{k}) . (- 10 \hat{i} - 47 \hat{j} + 39 \hat{k}) = - 30 + 312 = 282 We observe (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) \neq 0 Thus, the given lines do not intersect .$

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Similar questions

Q. Determine whether the following pair of lines intersect or not:
(i)

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

(ii)

\frac{x - 1}{2} = \frac{y + 1}{3} = z and \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2

(iii)

\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}

(iv)

\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}

Q. 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})

Q. Find the shortest distance between the following pairs of parallel lines whose equations are:
(i)

\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})

Q. The lines

¯ ¯ ¯ r = ˆ i - ˆ j + λ (2 ˆ i + ˆ k)

and

r = (2 ˆ i - ˆ j) + μ (ˆ i + ˆ j - ˆ k)

intersect for

Show that the lines $¯ ¯ ¯ r = (\land i + \land j - \land k) + λ (3 \land i - \land j)$ and $¯ ¯ ¯ r = (\land 4 i - \land k) + μ (2 \land i - 3 \land k)$ intersect. Find their point of intersection.

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By computing the shortest distance determine whether the following pairs of lines intersect or not: (i) r→=i^-j^+λ2i^+k^ and r→=2i^-j^+μi^+j^-k^ (ii) r→=i^+j^-k^+λ3i^-j^ and r→=4i^-k^+μ2i^+3k^ (iii) x-12=y+13=z and x+15=y-21; z=2 (iv) x-54=y-7-5=z+3-5 and x-87=y-71=z-53