Show that the lines x-3-3-y-11-z-55 and x-1-1-y-22-z-55 are coplanar- Hence- find the equation of the plane containing these lines-

The lines nbsp;x x1a1=y y1b1=z z1c1 nbsp;and nbsp;x x2a2=y y2b2=z z2c2 nbsp;are coplanar if nbsp;x2 x1y2 y1z2 z1a1b1c1a2b2c2=0. The given lines are ...

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Byju's Answer

Standard XII

Mathematics

Equation of Normal at a Point (x,y) in Terms of f'(x)

Show that the...

Question

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. Hence, find the equation of the plane containing these lines.

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Solution

The lines $\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}$ and $\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}$ are coplanar if $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| = 0$ .

The given lines are $\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}$ and $\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}$ .

Now, $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| =$ $|\begin{matrix} - 1 - (- 3) & 2 - 1 & 5 - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5 \end{matrix}| = |\begin{matrix} 2 & 1 & 0 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5 \end{matrix}| = 2 (5 - 10) - 1 (- 15 + 5) + 0 = - 10 + 10 + 0 = 0$

So, the given lines are coplanar.

The equation of the plane containing the given lines is

$|\begin{matrix} x - (- 3) & y - 1 & z - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5 \end{matrix}| = 0 \Rightarrow |\begin{matrix} x + 3 & y - 1 & z - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5 \end{matrix}| = 0 \Rightarrow (x + 3) (5 - 10) - (y - 1) (- 15 + 5) + (z - 5) (- 6 + 1) = 0 \Rightarrow - 5 (x + 3) + 10 (y - 1) - 5 (z - 5) = 0 \Rightarrow x - 2 y + z = 0$
Thus, the equation of the plane containing the given lines is x − 2y + z = 0.

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Similar questions

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. Show that the lines

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} and \frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

are coplanar. Also, find the equation of the plane containing them.

Q. If the lines

\frac{x - 1}{- 3} = \frac{y - 2}{- 2 k} = \frac{z - 3}{k}

and

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

are perpendicular, find the value of

k

and hence find the equation of plane containing these lines.

Q. Prove that the lines

\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}

and

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

are coplanar. Also, find the equation of the plane containing these two lines.

Q. If lines

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

and

\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}

intersect, then find the value of

k

and hence find the equation of the plane containing these lines.

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Show that the lines x+3-3=y-11=z-55 and x+1-1=y-22=z-55 are coplanar. Hence, find the equation of the plane containing these lines.

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. Hence, find the equation of the plane containing these lines.