Find the value of p so that the lines x-1 - -3 - y-p - 2 - z-2 - 1 and x - 1 - y-7 - -3 - z-7 - 2 are in the same plane-

The correct option is B 3 Let the dr' s of the required plane be a,b,c ⇒ 3a+2b+c=0 and a 3b+2c=0 ⇒ a / 4+3 = b / 1+6 = c / 9 2 ...

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Byju's Answer

Standard XII

Mathematics

Equation of a Plane Passing through Three Points

Find the valu...

Question

Find the value of $p$ so that the lines $\frac{x + 1}{- 3} = \frac{y - p}{2} = \frac{z + 2}{1}$ and $\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}$ are in the same plane.

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Solution

The correct option is B $3$
Let the dr' s of the required plane be $a, b, c$
$\Rightarrow - 3 a + 2 b + c = 0$ and $a - 3 b + 2 c = 0$
$\Rightarrow \frac{a}{4 + 3} = \frac{b}{1 + 6} = \frac{c}{9 - 2}$
$\Rightarrow \frac{a}{7} = \frac{b}{7} = \frac{c}{7}$
$\Rightarrow a = b = c$
Equation of plane passing through $(0, 7, - 7)$ and having dr' s $(1, 1, 1)$ is $1 (x - 0) + 1 (y - 7) + 1 (z + 7) = 0$
$\Rightarrow x + y + z = 0;$ It also contains point $(- 1, p, - 2)$
$\Rightarrow - 1 + p - 2 = 0$
$\Rightarrow p = 3$

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Find the value of p so that the lines x+1−3=y−p2=z+21 and x1=y−7−3=z+72 are in the same plane.

Find the value of $p$ so that the lines $\frac{x + 1}{- 3} = \frac{y - p}{2} = \frac{z + 2}{1}$ and $\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}$ are in the same plane.