Find the vector equation of the plane that passes through the point (1, 0, 0) and contains the line r = ⋏i Ģ‚ j Ģ‚

 

Find vector equation of plane that passes through the point (1,0,0)

Question 27 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 27 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3

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Transcript

Equation of plane passing through point A whose position vector is š’‚ āƒ— & perpendicular to š’ āƒ— is (š’“ āƒ— āˆ’ š’‚ āƒ—) . š’ āƒ— = 0 Given Plane passes through (1, 0, 0) So š’‚ āƒ— = 1š‘– Ģ‚ + 0š‘— Ģ‚ āˆ’ 0š‘˜ Ģ‚ š’‚ āƒ— = š‘– Ģ‚ Thus, equation of plane is (š‘Ÿ āƒ— āˆ’ š‘Ž āƒ—) . š‘› āƒ— = 0 (š’“ āƒ— āˆ’ š’Š Ģ‚) . š’ āƒ— = 0 Now, The plane contains the line š‘Ÿ āƒ—= šœ†š‘— Ģ‚ So, Line is perpendicular to normal of plane š’‹ Ģ‚ . š’ āƒ— = 0 And, Lines point (0, 0, 0) lies on plane as well Putting š‘Ÿ āƒ— = 0 āƒ— in equation of plane (0 āƒ— āˆ’ š‘– Ģ‚) . š‘› āƒ— = 0 āˆ’ š‘– Ģ‚ . š‘› āƒ— = 0 š’Š Ģ‚ . š’ āƒ— = 0 Since š’ āƒ— perpendicular to both š’Š Ģ‚ and š’‹ Ģ‚ Thus, š’ āƒ— = š’Œ Ģ‚ So, our equation of plane becomes (š‘Ÿ āƒ— āˆ’ š‘– Ģ‚) . š‘› āƒ— = 0 (š’“ āƒ— āˆ’ š’Š Ģ‚) . š’Œ Ģ‚ = 0 š‘Ÿ āƒ— . š‘˜ Ģ‚ āˆ’ š‘– Ģ‚ . š‘˜ Ģ‚ = 0 š’“ āƒ— . š’Œ Ģ‚ = 0

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