Example 14 - Find direction cosines of unit vector perpendicular - Examples

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  1. Chapter 11 Class 12 Three Dimensional Geometry
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Example 14 Find the direction cosines of the unit vector perpendicular to the plane š‘Ÿļ·Æ .(6 š‘–ļ·Æ āˆ’ 3 š‘—ļ·Æ ā€“ 2 š‘˜ļ·Æ) + 1 = 0 passing through the origin. Vector equation of a plane at a distance ā€˜dā€™ from the origin and unit vector to normal from origin š‘›ļ·Æ is š’“ļ·Æ . š’ļ·Æ = d Unit vector of š‘›ļ·Æ = š‘›ļ·Æ = 1ļ·® š‘›ļ·Æļ·Æļ·Æ( š‘›ļ·Æ) Given, equation of plane is š‘Ÿļ·Æ.(6 š‘–ļ·Æ āˆ’ 3 š‘—ļ·Æ āˆ’ 2 š‘˜ļ·Æ) + 1 = 0 š‘Ÿļ·Æ.(6 š‘–ļ·Æ āˆ’ 3 š‘—ļ·Æ āˆ’ 2 š‘˜ļ·Æ) = āˆ’ 1 Multiplying with āˆ’1 on both sides, āˆ’ š‘Ÿļ·Æ.(6 š‘–ļ·Æ āˆ’ 3 š‘—ļ·Æ āˆ’ 2 š‘˜ļ·Æ) = āˆ’ 1 Ɨ āˆ’1 š’“ļ·Æ. (ā€“6 š’Šļ·Æ + 3 š’‹ļ·Æ + 2 š’Œļ·Æ) = 1 So; š‘›ļ·Æ = āˆ’ 6 š‘–ļ·Æ + 3 š‘—ļ·Æ + 2 š‘˜ļ·Æ Magnitude of š‘›ļ·Æ = ļ·® āˆ’ 6ļ·Æ2+32+22ļ·Æ š‘›ļ·Æļ·Æ = ļ·®36+9+4ļ·Æ = ļ·®49ļ·Æ = 7 Now, š’ļ·Æ = 1ļ·® š‘›ļ·Æļ·Æļ·Æ ( š‘›ļ·Æ) = 1ļ·®7ļ·Æ (āˆ’ 6 š‘–ļ·Æ + 3 š‘—ļ·Æ + 2 š‘˜ļ·Æ) = āˆ’šŸ”ļ·®šŸ•ļ·Æ š’Šļ·Æ + šŸ‘ļ·®šŸ•ļ·Æ š’‹ļ·Æ + šŸļ·®šŸ•ļ·Æ š’Œļ·Æ āˆ“ Direction cosines of unit vector perpendicular to the given plane i.e. in are āˆ’ šŸ”ļ·®šŸ•ļ·Æ, šŸ‘ļ·®šŸ•ļ·Æ , šŸļ·®šŸ•ļ·Æ .

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