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Find the equation of the plane containing two parallel lines $\mathrm { \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{3}}$ and $\mathrm { \frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{6}}$ . Also, find if the plane thus obtained contains the line $\mathrm { \frac{x}{4} = \frac{y-2}{-2} = \frac{z+1}{6}}$ or not.

Answers (1)

Let $\mathrm { L_1 : \ \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{3}}$ and $\mathrm { L_2: \frac{x}{4} = \frac{y-2}{-2} = \frac{z+1}{6}}$ .

The points in these lines are (1,-1,0) & (0,2,-1) repectively.

Let the d.r's of the normal to the required plane be A, B, C. So the equation of plane is.

$\mathrm{A(x-1) + B(y+1) + C(z - 0) = 0\quad - (i)}$

As (0,2,-1) lies on (i), so, $\mathrm{-A + 3B - C = 0\quad - (ii)}$

Also as the plane contains the line so, the normal to the plane shall be $\mathrm{\perp}$ to these lines also.

i.e $\mathrm{2A -B +3 C = 0\quad - (iii)}$ (as d.r's of the line L₁ are 2,-1,3)

Solving (ii) and (iii) we get: $\mathrm{\frac{A}{8} = \frac{B}{1} = \frac{C}{-5}}$

By (i), we have $\mathrm{8(x-1) + 1(y+1) -5(z - 0) = 0}$

i.e $\mathrm{8x + y - 5z = 7}$

Now let $\mathrm {L_3: \frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{6}}$ . Clearly the point on it is M(2,1,2) and its d.r's are 3,1,5

$\because$ M satisfies the plane $\mathrm{8x + y - 5z = 7}$ as

$\mathrm{LHS : 8x + y - 5z = 8\times 2 + 1 - 5\times 2 = 7 = RHS}$ .

And $\mathrm{8\times 3 + 1\times 1 - 5\times5 = 24 + - 25 = 0}$ i.e normal of the plane is $\mathrm{\perp}$ to the line L₃ also.

Hence, the plane $\mathrm{8x + y - 5z = 7}$ contains the line L₃.

Posted by

Ravindra Pindel

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