Show that the lines

are intersecting. Hence find their point of intersection.

OR

Find the vector equation of the plane through the points (2, 1, – 1) and (– 1, 3, 4) and perpendicular to the plane x – 2y + 4z =10.

Given equation of lines are -

∴ equations of line can be written in cartesian form as –

Random coordinates of this line is given by –

x = λ + 3 ; y = 2λ + 2 and z = 2λ - 4

And other equation is-

Random coordinates of this line is given by –

x = 3μ + 5 ; y = 2μ - 2 and z = 6μ

for intersection of these two lines we should have –

λ + 3 = 3μ + 5 …(1)

2λ + 2 = 2μ - 2 …(2)

2λ – 4 = 6μ …(3)

Solving equation 1 and 2 we get-

μ = -2 and λ = -4

Clearly this value also satisfies equation 3, so lines will intersect and point of intersection can be given by putting values of λ or μ in their respective random coordinates.

So by putting the value of μ we get coordinate of intersection is

(-1, -6, -12)

OR

As we are given with the Position vector of two points say P and Q.

Given, and

As P and Q lie on the given plane.

∴ is a vector on the given plane.

⇒

∴

To write the equation now we need a unit vector perpendicular to the given plane.

Given that plane is perpendicular to another plane x – 2y + 4z =10.

A vector perpendicular to this plane is

∴ unit vector of this plane will be parallel to the given plane.

To find the unit vector perpendicular to our plane we need to take cross product of

∴ =

Expanding the determinant about first row:

⇒

⇒

∴ equation of plane is given as –

⇒ ((

⇒

⇒

Or, 18x + 17y + 14z = 0 is the required equation of plane.