Misc 11 - Find coordinates of point where line through (5, 1, 6) and

Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5 Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

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Question 7 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.The equation of a line passing through two points with position vectors š‘Ž āƒ— & š‘ āƒ— is š’“ āƒ— = š’‚ āƒ— + šœ†(š’ƒ āƒ— āˆ’ š’‚ āƒ—) Given, the line passes through (š‘ āƒ— āˆ’ š‘Ž āƒ—) = (3š‘– Ģ‚ + 4š‘— Ģ‚ + 1š‘˜ Ģ‚) āˆ’ (5š‘– Ģ‚ + 1š‘— Ģ‚ + 6š‘˜ Ģ‚) = (3 āˆ’5)š‘– Ģ‚ + (4 āˆ’ 1)š‘— Ģ‚ + (1 āˆ’ 6)š‘˜ Ģ‚ A (5, 1, 6) š‘Ž āƒ— = 5š‘– Ģ‚ + 1š‘— Ģ‚ + 6š‘˜ Ģ‚ B(3, 4, 1) š‘ āƒ— = 3š‘– Ģ‚ + 4š‘— Ģ‚ + 1š‘˜ Ģ‚ = āˆ’2š‘– Ģ‚ + 3š‘— Ģ‚ āˆ’ 5š‘˜ Ģ‚ ∓ š’“ āƒ— = (5š’Š Ģ‚ + š’‹ Ģ‚ + 6š’Œ Ģ‚) + šœ† (āˆ’2š’Š Ģ‚ + 3š’‹ Ģ‚ āˆ’ 5š’Œ Ģ‚) Let the coordinates of the point where the line crosses the ZX plane be (x, 0, z) So, š’“ āƒ— = xš’Š Ģ‚ + 0š’‹ Ģ‚ + zš’Œ Ģ‚ Since point lies in line, it will satisfy its equation, Putting (2) in (1) xš‘– Ģ‚ + 0š‘— Ģ‚ + zš‘˜ Ģ‚ = 5š‘– Ģ‚ + š‘— Ģ‚ + 6š‘˜ Ģ‚ āˆ’2šœ†š‘– Ģ‚ + 3šœ†š‘— Ģ‚ āˆ’ 5šœ†š‘˜ Ģ‚ xš‘– Ģ‚ + 0š‘— Ģ‚ + zš‘˜ Ģ‚ = (5 āˆ’2šœ†)š‘– Ģ‚ + (1 + 3šœ†)š‘— Ģ‚ + (6 āˆ’ 5šœ†)š‘˜ Ģ‚ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 3šœ† 3šœ† = āˆ’1 ∓ šœ† = (āˆ’šŸ)/šŸ‘ Now, x = 5 āˆ’ 2šœ† = 5 āˆ’ 2 Ɨ (āˆ’1)/3 = 5 + 2/3 = 17/13 z = 6 āˆ’ 5šœ† = 6 āˆ’ 5 Ɨ (āˆ’1)/3 = 6 + 5/3 = 23/3 Therefore, the coordinate of the required point are (šŸšŸ•/šŸ‘,šŸŽ,šŸšŸ‘/šŸ‘) Question 7 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.The equation of a line passing through two points A(š‘„_1, š‘¦_1, š‘§_1) and B(š‘„_2, š‘¦_2, š‘§_2) is (š’™ āˆ’ š’™_šŸ)/(š’™_šŸ āˆ’ š’™_šŸ ) = (š’š āˆ’ š’š_šŸ)/(š’š_šŸ āˆ’ š’š_šŸ ) = (š’› āˆ’ š’›_šŸ)/(š’›_šŸ āˆ’ š’›_šŸ ) Given the line passes through the points A (5, 1, 6) ∓ š‘„_1= 5, š‘¦_1= 1, š‘§_1= 6 B(3, 4, 1) ∓ š‘„_2= 3, š‘¦_2= 4, š‘§_2= 1 So, the equation of line is (š‘„ āˆ’ 5)/(3 āˆ’ 5) = (š‘¦ āˆ’ 1)/(4 āˆ’ 1) = (š‘§ āˆ’ 6)/(1 āˆ’ 6) (š’™ āˆ’ šŸ“)/(āˆ’šŸ) = (š’š āˆ’ šŸ)/šŸ‘ = (š’› āˆ’ šŸ”)/(āˆ’šŸ“) = k So, Since the line crosses the ZX plane at (x, 0, z), y = 0 3k + 1 = 0 3k = āˆ’1 ∓ k = (āˆ’šŸ)/šŸ‘ So, x = –2k + 5 = āˆ’2 Ɨ (āˆ’1)/3 + 5 = 2/3 + 5 = 17/3 y = 0 & z = āˆ’5k + 6 = āˆ’5 Ɨ (āˆ’1)/3 + 6 = 5/3 + 6 = 23/3 therefore, the coordinate of the required point are (šŸšŸ•/šŸ‘,šŸŽ,šŸšŸ‘/šŸ‘)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo