Q. 215.0( 2 Votes )

# Show that the lin

Answer :

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.

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