Question

Show that the line of intersection of the planes

x + 2 y + 3 z = 8

and

2 x + 3 y + 4 z = 11

is coplanar with the line

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

. Also find the equation of the plane containing them.

Accepted Answer

The given line $\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z + 1}{3} . . . . . (1)$ is coplanar with the line determined by the planes $x + 2 y + 3 z - 8 = 0 \dots (2)$ and $2 x + 3 y + 4 z - 11 = 0 \dots (3)$ , if we are able to show there exists a plane passing through intersection of planes $(2)$ and $(3)$ containing the line $(1)$ .

Equation of the plane passing through the intersection of planes $(2)$ and $(3)$ is $(x + 2 y + 3 z - 8) + k (2 x + 3 y + 4 z - 11) = 0 . . . . . (4)$

We find, value of k for which the plane given by $(4)$ passes through the point $(- 1, - 1, - 1)$ lying on line $(1)$ .

Substituting the coordinates of the point $(- 1, - 1, - 1)$ in $(4)$ , we get

$(- 1 - 2 - 3 - 8) + k (- 2 - 3 - 4 - 11) = 0$

$\Rightarrow - 14 - 20 k = 0$

$\Rightarrow k = \frac{- 7}{10}$

Putting, $k = \frac{- 7}{10}$ in $(4)$ , we get

$(x + 2 y + 3 z - 8) - \frac{7}{10} (2 x + 3 y + 4 z - 11) = 0$

$\Rightarrow 4 x + y - 2 z + 3 = 0 . . . . . (5)$

Now, we find the value of $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 1 (4) + 2 (1) + 3 (- 2)$

i.e., $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

Which, implies line $(1)$ lies in plane $(5)$

Hence, the two lines are coplanar and the equation of the plane containing them is $4 x + y - 2 z + 3 = 0$

Odisha Board Solutions for Chapter: Three Dimensional Geometry, Exercise 3: EXERCISES -13(C)

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Three Dimensional Geometry, Exercise 3: EXERCISES -13(C)

Questions from Odisha Board Solutions for Chapter: Three Dimensional Geometry, Exercise 3: EXERCISES -13(C) with Hints & Solutions

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