Find the equation of the line of intersection of the two plane given by x+y+z=5 and 4x+y+2z=15

Pls help me to solve this.. After i use cross product i got (1,2,-3)..Then what should i do?

To find the equation of the line of intersection of two planes, you need to find a vector that is orthogonal (perpendicular) to both planes. This can be done by taking the cross product between the normal vectors of the two planes.

First, let's find the normal vectors of the two planes:

The equation of the first plane is x + y + z = 5.
To find the normal vector of this plane, we can extract the coefficients of x, y, and z: (1, 1, 1).

The equation of the second plane is 4x + y + 2z = 15.
Again, we extract the coefficients of x, y, and z to get the normal vector: (4, 1, 2).

Now, take the cross product of these two normal vectors:

(1, 1, 1) × (4, 1, 2)
= ((1)(2) - (1)(1), (1)(4) - (1)(2), (1)(1) - (1)(4))
= (1, 2, -3)

You have correctly calculated the cross product as (1, 2, -3). This vector is orthogonal to both planes.

To write the equation of the line of intersection, we need a point on the line. We can choose any point that satisfies both plane equations. Let's solve the system of equations:

x + y + z = 5
4x + y + 2z = 15

Multiply the first plane equation by 4:
4(x + y + z) = 4(5)
4x + 4y + 4z = 20

Subtract the second plane equation from this:
4x + 4y + 4z - (4x + y + 2z) = 20 - 15
4x + 4y + 4z - 4x - y - 2z = 5
3y + 2z = 5

Choose values for y and z to solve for x:
Let y = 0, then 2z = 5, giving z = 5/2.
Let z = 0, then 3y = 5, giving y = 5/3.

Therefore, one point on the line of intersection is (x, y, z) = (0, 5/3, 0).

Finally, write the equation of the line using the point (0, 5/3, 0) and the direction vector (1, 2, -3):

x = 0 + t(1)
y = 5/3 + t(2)
z = 0 + t(-3)

where t is a scalar parameter.

Thus, the equation of the line of intersection is:
x = t
y = 5/3 + 2t
z = -3t