Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Question 10 Find the vector equation of the plane passing through the intersection of the planes ๐ โ . (๐ ฬ + ๐ ฬ + ๐ ฬ) = 6 and ๐ โ . (2๐ ฬ + 3๐ ฬ + 4๐ ฬ) = โ 5, and the point (1, 1, 1).The vector equation of a plane passing through the intersection of planes ๐ โ. (๐1) โ = d1 and ๐ โ. (๐2) โ = d2 and also through the point (x1, y1, z1) is ๐ โ.((๐๐) โ + ๐(๐๐) โ) = d1 + ๐d2 Given, the plane passes through ๐ โ.(๐ ฬ + ๐ ฬ + ๐ ฬ) = 6 Comparing with ๐ โ.(๐1) โ = d1, (๐๐) โ = ๐ ฬ + ๐ ฬ + ๐ ฬ & d1 = 6 ๐ โ.(2๐ ฬ + 3๐ ฬ + 4๐ ฬ) = โ5 โ๐ โ.(2๐ ฬ + 3๐ ฬ + 4๐ ฬ) = 5 ๐ โ .(โ 2๐ ฬ โ 3๐ ฬ โ 4๐ ฬ) = 5 Comparing with ๐ โ.(๐2) โ = d2 (๐๐) โ = โ 2๐ ฬ โ 3๐ ฬ โ 4๐ ฬ & d2 = 5 Equation of plane is ๐ โ. [(๐ ฬ+๐ ฬ+๐ ฬ )+"๐" (โ2๐ ฬโ3๐ ฬโ4๐ ฬ)] = 6 + ๐5 ๐ โ. [(๐ ฬ" " +๐ ฬ" " +๐ ฬ )โ"๐" (๐๐ ฬ+๐๐ ฬ+๐๐ ฬ)] = 6 + 5๐ Now to find ๐ , put ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ (x๐ ฬ + y๐ ฬ + z๐ ฬ). [(๐ ฬ+๐ ฬ+๐ ฬ )โ"๐" (2๐ ฬ+3๐ ฬ+4๐ ฬ)] = 5๐ + 6 (x๐ ฬ + y๐ ฬ + z๐ ฬ).(๐ ฬ+๐ ฬ+๐ ฬ ) โ ๐ (x๐ ฬ + y๐ ฬ + z๐ ฬ).(2๐ ฬ+3๐ ฬ+4๐ ฬ) = 5๐ + 6 (x ร 1) + (y ร 1) + (z ร 1) โ ๐[(๐ฅร2)+(๐ฆร3)+(๐งร4)] = 5๐ + 6 x + y + z โ ๐[2๐ฅ+3๐ฆ+4๐ง] = 5๐ + 6 x + y + z โ 2๐๐ฅ โ 3๐y โ 4๐z = 5๐ + 6 (1 โ 2๐)x + (1 โ 3๐)y + (1 โ 4๐) z = 5๐ + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 โ 2๐)x + (1 โ 3๐)y + (1 โ 4๐) z = 5๐ + 6 (1 โ2๐) ร 1 + (1 โ 3๐) ร 1 + (1 โ 4๐) ร 1 = 5๐ + 6 1 โ2๐ + 1 โ 3๐ + 1 โ 4๐= 5๐ + 6 3 โ 9๐ = 5๐ + 6 โ14๐ = 3 โด ๐ = (โ๐)/๐๐ Putting value of ๐ in (1), ๐ โ. [(๐ ฬ" " +" " ๐ ฬ" " +" " ๐ ฬ )โ(( โ3)/14)(2๐ ฬ+3๐ ฬ+"4" ๐ ฬ)]= 6 + 5 ร ( โ3)/14 ๐ โ. [(๐ ฬ+๐ ฬ+" " ๐ ฬ )+3/14(2๐ ฬ+3๐ ฬ+"4" ๐ ฬ)]= 6 โ 15/14 ๐ โ. [๐ ฬ+๐ ฬ" " +๐ ฬ+ 6/14 ๐ ฬ+9/14 ๐ ฬ+12/14 ๐ ฬ ]= 69/14 ๐ โ. [(1+6/14) ๐ ฬ +(1+9/14) ๐ ฬ+(1+12/14) ๐ ฬ ]= 69/14 ๐ โ. [20/14 ๐ ฬ + 23/14 ๐ ฬ + 26/14 ๐ ฬ ]= 69/14 ๐ โ. [1/14(20๐ ฬ+23๐ ฬ+26๐ ฬ)]= 69/14 1/14 ๐ โ. (20๐ ฬ + 23๐ ฬ + 26๐ ฬ) = 69/14 ๐ โ. (20๐ ฬ + 23๐ ฬ + 26๐ ฬ) = 69 Therefore, the vector equation of the required plane is ๐ โ.(๐๐๐ ฬ + ๐๐๐ ฬ + ๐๐๐ ฬ) = ๐๐