The equation of the plane containing the two lines x-12-y-1-1-z-03 and x-2-y-2-3-z-1-1 isa 8 x-y-5 z-7-0b 8 x-y-5 z-7-0c 8 x-y-5 z-7-0d None of these

(d) None of these x #8722;12=y+1 #8722;1=z #8722;03 nbsp;and nbsp; x #8722;2=y #8722;2 #8722;3=z+1 #8722;1 Now, if these two lines lie on a ...

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Standard XII

Mathematics

Cramer's Rule

The equation ...

Question

The equation of the plane containing the two lines $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} and \frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}$ is
(a) 8x + y − 5z − 7 = 0
(b) 8x + y + 5z − 7 = 0
(c) 8x − y − 5z − 7 = 0
(d) None of these

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Solution

(d) None of these

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

Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.

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The equation of the plane containing the two lines x-12=y+1-1=z-03 and x-2=y-2-3=z+1-1 is (a) 8x + y − 5z − 7 = 0 (b) 8x + y + 5z − 7 = 0 (c) 8x − y − 5z − 7 = 0 (d) None of these

(d) None of these x−12=y+1−1=z−03 and x−2=y−2−3=z+1−1 Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.

The equation of the plane containing the two lines $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} and \frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}$ is
(a) 8x + y − 5z − 7 = 0
(b) 8x + y + 5z − 7 = 0
(c) 8x − y − 5z − 7 = 0
(d) None of these

(d) None of these

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

Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.