Answer :

**Given:** A plane passing through the points A(2,1, 2) and B(4, -2, 1) and perpendicular to plane

**To Find:** Equation of the plane. Also find coordinates of the point where the line passing through the points (3,4,1) and (5,1,6) crosses the plane thus obtained.

Let, P(x,y,z) be a point on the plane.

A and B both lie on the same plane.

Normal to the plane is =

And this plane is perpendicular to the plane

Normal of this plane is,

So, we can say,

--- (i)

Here, and

Putting the values in (i) we get,

-(y – 1) + 3(z – 2) -2{-3(x – 2) – 2(y – 1)} = 0

-y + 1 + 3z – 6 + 6(x – 2) + 4(y – 1) = 0

-y + 1 + 3z – 6 + 6x – 12 + 4y – 4 = 0

6x + 3y + 3z = 21

2x + y + z = 7

So, the equation of the plane is 2x + y + z = 7 **[Answer(i)]**

Equation of line passing through the points (3,4,1) and (5,1,6) is,

So, x = 2k + 3

y = -3k + 4

z = 5k + 1

Putting the values of x, y, z in the previous plane equation we get,

2(2k + 3) – 3k + 4 + 5k + 1 = 7

4k + 6 – 3k + 4 + 5k + 1 = 7

6k = -4

k =

So, the coordinates of the point crossing the plane are,

x = 2 + 3 =

y = -3 + 4 = 6

z = 5 + 1 =

So, the point is = (, 6, ) **[Answer(ii)]**

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