Example 15  - Find distance of plane from origin - Class 12 - Examples

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Example 15 (Method 1) Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin. Given, the equation of plane is 2x − 3y + 4z − 6 = 0 2x − 3y + 4z = 6 Direction ratios of 𝑛﷯ = 𝑎, 𝑏, 𝑐 a = 2, b = −3, c = 4 Also, ﷮ 𝑎﷮2﷯+ 𝑏﷮2﷯+ 𝑐﷮2﷯﷯ = ﷮ 2﷮2﷯ + (− 3)﷮2﷯ + 4﷮2﷯﷯ = ﷮4+9+16﷯ = ﷮29﷯ ∴ Direction cosines are l = 𝑎﷮ ﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ + 𝑐﷮2﷯﷯﷯ , m = 𝑏﷮ ﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ + 𝑐﷮2﷯﷯﷯ , n = 𝑐﷮ ﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ + 𝑐﷮2﷯﷯﷯ l = 2﷮ ﷮29﷯﷯ , m = − 3﷮ ﷮29﷯﷯ ,n = 4﷮ ﷮29﷯﷯ Equation of plane is lx + my + nz = d 2﷮ ﷮29﷯﷯ x – 3﷮ ﷮29﷯﷯ y + 4﷮ ﷮29﷯﷯ z = d 2x − 3y + 4z = d ﷮29﷯ Comparing with (1) i.e. 2x − 3y + 4z = 6, d ﷮29﷯ = 6 d = 𝟔﷮ ﷮𝟐𝟗﷯﷯ Example 15 (Method 2) Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin. Distance of point P(x1, y1, z1) from plane Ax + By + Cz = D is d = 𝐴 𝑥﷮1﷯ + 𝐵 𝑦﷮1﷯ + 𝐶 𝑧﷮1﷯ − 𝐷﷮ ﷮ 𝐴﷮2﷯ + 𝐵﷮2﷯ + 𝐶﷮2﷯﷯﷯﷯ Since we have to find distance from Origin P(x1, y1, z1) = O(0, 0, 0) ∴ x1 = 0, y1 = 0, z1 = 0 & plane is 2x – 3y + 4z – 6 = 0 2x – 3y + 4z = 6 Comparing with Ax + By + Cz = D A = 2, B = –3, C = 4 & D = 6 Putting values in formula d = 𝐴 𝑥﷮1﷯ + 𝐵 𝑦﷮1﷯ + 𝐶 𝑧﷮1﷯ − 𝐷﷮ ﷮ 𝐴﷮2﷯ + 𝐵﷮2﷯ + 𝐶﷮2﷯﷯﷯﷯ d = 2 0﷯ − 3 0﷯ + 4 0﷯ − 6﷮ ﷮ 2﷮2﷯ + (−3)﷮2﷯ + 4﷮2﷯﷯﷯﷯ d = −6﷮ ﷮4 + 9 + 16﷯﷯﷯ d = −6﷮ ﷮29﷯﷯﷯ d = 𝟔﷮ ﷮𝟐𝟗﷯﷯

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