Show that the lines 5-x-4-y-74-z-3-5 and x -87-2 y -82- z -53 are coplanar- -CBSE 2014-

The equations of the given lines can be re written as x 54=y 74=z+3 5 and x 87=y 41=z 53 We know that the lines x x1l1=y y1m1=z z1n1 and x x2l2=y y2m2=z z2n ...

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Byju's Answer

Standard XII

Mathematics

Asymptotes

Show that the...

Question

Show that the lines $\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}$ and $\frac{x - 8}{7} = \frac{2 y - 8}{2} = \frac{z - 5}{3}$ are coplanar. [CBSE 2014]

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Solution

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

$\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}$

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

We know that the lines $\frac{x - x_{1}}{l_{1}} = \frac{y - y_{1}}{m_{1}} = \frac{z - z_{1}}{n_{1}}$ and $\frac{x - x_{2}}{l_{2}} = \frac{y - y_{2}}{m_{2}} = \frac{z - z_{2}}{n_{2}}$ are coplanar if $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{matrix}| = 0$ .

Here,

$x_{1} = 5, y_{1} = 7, z_{1} = - 3, x_{2} = 8, y_{2} = 4, z_{2} = 5 l_{1} = 4, m_{1} = 4, n_{1} = - 5, l_{2} = 7, m_{2} = 1, n_{2} = 3$

$∴ |\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{matrix}| = |\begin{matrix} 8 - 5 & 4 - 7 & 5 - (- 3) \\ 4 & 4 & - 5 \\ 7 & 1 & 3 \end{matrix}| = |\begin{matrix} 3 & - 3 & 8 \\ 4 & 4 & - 5 \\ 7 & 1 & 3 \end{matrix}| = 3 (12 + 5) + 3 (12 + 35) + 8 (4 - 28) = 51 + 141 - 192 = 0$

So, the given lines are coplanar.

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Similar questions

Q. Show that the lines

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

and

\frac{x - 8}{7} = \frac{2 y - 8}{2} = \frac{z - 5}{3}

are coplanar.

Q. Show that the lines

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

and

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

are coplanar.

Q. show that the lines

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

and

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

are coplanar. Also find the equation of the plane.

Q. Show that the lines

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

are coplanar. Also find the equation of plane containing the lines.

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}

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Show that the lines 5-x-4=y-74=z+3-5 and x-87=2y-82=z-53 are coplanar. [CBSE 2014]

Show that the lines $\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}$ and $\frac{x - 8}{7} = \frac{2 y - 8}{2} = \frac{z - 5}{3}$ are coplanar. [CBSE 2014]