Question

Find the point at which the line x = 5 − t, y = 2 + t, z = 3t intersects the plane x − y + 4z = 13.

Answer 1

The value of the points at which intersection of plane takes place is (x,y,z) = (4,3,3).

According to the statement

we have to find the point which they intersect the given plane.

So, For this purpose, we know that the

The given equations is :

x = 5 − t, -(1)

y = 2 + t,  -(2)

z = 3t -(3)

And the given plane is :

x − y + 4z = 13  -(4)

And, the intersection of plane point is :

Put the equations 1 and 2 and 3 in the 4th equations then

(5 − t) - (2 + t) + 4(3t) = 13

3 - 2t + 12t = 13

3 + 10t = 13

10t = 10

Then

t = 1

Put these values in 1 and 2 and 3rd equation then the value of x y z become

x = 5 − t, -(1)

x = 5 -1

x = 4

And

y = 2 + t,  -(2)

y = 2 + 1

y = 3

And

z = 3t -(3)

z = 3(1)

z = 3.

The value of (x,y,z) = (4,3,3).

So, The value of the points at which intersection of plane takes place is (x,y,z) = (4,3,3).

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