Q. 235.0( 1 Vote )

# Find the distance

Given Data:

Plane 3x + y – z + 2 = 0

Equation of line passing through (3,–2, 1) and parallel to the line

Any general point on this line is (2α+3,–3α–2, α+1)

As line will intersect plane 3x + y – z + 2 = 0 at some point,

3(2α+3)–3α–2 – α–1 + 2 = 0

α = –4

Therefore this point is (–5, 10,–3)

Distance between the points,

d = 4√14

Now the line passing through (3,–2, 1) and perpendicular to the plane 3x + y – z + 2 = 0 is

Any general point on this line is (3β+3, β–2, –β+1)

If it pass through the plane then it should suffice the plane equation

3(3β+3) + β–2 + β –1 + 2 = 0

Putting this value in (3β+3, β–2,–β+1) we get

OR

Given Data:

Point: (3, –2, –1)

Planes:

The intersection line of two planes is

If this line passes through the point (3, –2, –1)

(2+α)3 – 2(3+α) – 1(–1–2α) + 1=0

Therefore equation of plane is

The angle between the two given planes is

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