Find the equation of the plane through the line

and parallel to the line

Hence, find the shortest distance between the lines.

OR

Show that the line of intersection of the planes x+2y+3z=8 and 2x+3y+4z=11 is coplanar with the line

Also, find the equation of the plane containing them.

The two given lines are:

Let a, b, c be the direction Ratios of the normal to the plane containing the line (1).

Also, as plane contains line (1), so it passes through (1, 4, 4)

Therefore, equation of plane is -

a (x –1) + b(y – 4)+c(z – 4) = 0 …(3)

We know that if a₁, b₁, c₁ are direction Ratios of a plane and a₂, b₂, c₂ are direction Ratios of a line parallel to it then,

a₁a₂ + b₁b₂ + c₁c₂ = 0

As, required plane is parallel to line (1),

⇒ 3a+2b – 2c = 0 … (4)

Also, as required plane is parallel to line (2),

⇒ 2a – 4b + 1c = 0 … (5)

Solving (4) and (5),

Let,

Putting, a = 6λ, b = λ7, c = 16λ in (3), we get

⇒ 6λ(x –1) + 7λ(y – 4) + 16λ(z – 4) = 0

⇒ λ((6x –6) + (7y – 28) + (16z – 64)) = 0

⇒ 6x + 7y +16z – 6 – 28 – 64 = 0,

⇒ 6x + 7y +16z – 98 = 0,

which is the required equation of the plane.

Since line (2) is parallel to required plane

∴ SD between two lines = Perpendicular distance of the point

(–1, 1 –2) from the plane.

Distance between a point (x’, y’, z’) from a line

ax + by + cz – d = 0 is -

Therefore,

OR

Equation of given line,

Also, (1) is coplanar with the line determined by the planes

x + 2y + 3z – 8 = 0 … (2)

and

2x + 3y + 4z – 11 = 0 … (3)

So, need to show there exists a plane passing through intersection of planes (2) and (3) containing the line (1).

Any plane passing through intersection of two planes P1 and P2 is

P1 + kP2 = 0

Where k is some constant,

So, equation of the plane passing through the intersection of planes (2) and (3) is -

(x + 2y + 3z – 8) + k (2x + 3y + 4z – 11) = 0 … (4)

As, (4) passes through (-1, -1, -1) so, substituting the coordinates of the point (–1, –1, –1) in (4), we get,

(–1–2–3–8) + k (–2–3–4–11) = 0

⇒ (-14) + k (-20) = 0

Putting in (4) we get,

⇒ 10(x + 2y + 3z – 8) -7(2x + 3y + 4z – 11) = 0

⇒ 10x + 20y + 30z – 80 – 14x + 21y + 28z – 77 = 0

⇒ 4x + y – 2z + 3 = 0

So, plane passing through intersection of planes (2) and (3) is-

4x + y – 2z + 3 =0 … (5)

Now we find value of a₁a₂ + b₁b₂ + c₁c₂ where a₁, b₁, c₁ are

Direction Ratios of the line (1) and a₂, b₂, c₂ are Direction Ratios of the normal to the plane (5)

From equation (1) – (a₁, b₁, c₁) = (1, 2, 3) and

From equation (5) – (a₂, b₂, c₂) = (4, 1, -2)

So, a₁a₂ + b₁b₂ + c₁c₂ = (1)(4) + (2)(1) + (3) (-2)

= 4 + 2 – 6

= 0

i.e., a₁a₂ + b₁b₂ + c₁c₂ = 0

which implies line (1) lies in plane (5)

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