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Example 24 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at March 22, 2023 by

Example 24 - Find distance of point (2, 5, -3) from plane - Examples

Example 24 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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Example 24 Find the distance of a point (2, 5, –3) from the plane 𝑟﷯ . (6 𝑖﷯ – 3 𝑗﷯ + 2 𝑘﷯) = 4 The distance of a point with position vector 𝑎﷯ from the plane 𝑟﷯. 𝑛﷯ = d, where 𝑛﷯ is the normal to the plane is 𝒂﷯. 𝒏﷯ − 𝒅﷮ 𝒏﷯﷯﷯﷯ Given, the point is (2, 5, −3) So, 𝑎﷯ = 2 𝑖﷯ + 5 𝑗﷯ − 3 𝑘﷯ The equation of plane is 𝑟﷯.(6 𝑖﷯ − 3 𝑗﷯ + 2 𝑘﷯) = 4 Comparing with 𝑟﷯. 𝑛﷯ = d, 𝑛﷯ = 6 𝑖﷯ − 3 𝑗﷯ + 2 𝑘﷯ & d = 4 Distance of point from plane = 𝑎﷯. 𝑛﷯ − 𝑑﷮ 𝑛﷯﷯﷯﷯ = 2 𝑖﷯ + 5 𝑗﷯ − 3 𝑘﷯﷯. 6 𝑖﷯ − 3 𝑗﷯ + 2 𝑘﷯﷯ − 4﷮ ﷮ 6﷮2﷯ + −3﷯﷮2﷯ + 2﷮2﷯﷯﷯﷯ = 2 × 6﷯ + 5 × −3﷯ + −3 × 2﷯ − 4﷮ ﷮36 + 9 + 4﷯﷯﷯ = 12 − 15 − 6 − 4﷮ ﷮49﷯﷯﷯ = −13﷮7﷯﷯ = 𝟏𝟑﷮𝟕﷯

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