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Misc 19 - Find vector equation of line passing through (1, 2, 3) and

Misc 19 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 19 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 19 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 19 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5
Misc 19 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6
Misc 19 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7


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Misc 19 (Method 1) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes š‘Ÿ āƒ— . (š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 2 š‘˜ Ģ‚) = 5 and š‘Ÿ āƒ— . (3š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚) = 6 . The vector equation of a line passing through a point with position vector š‘Ž āƒ— and parallel to a vector š‘ āƒ— is š’“ āƒ— = š’‚ āƒ— + šœ†š’ƒ āƒ— Given, the line passes through (1, 2, 3) So, š‘Ž āƒ— = 1š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ Given, line is parallel to both planes āˆ“ Line is perpendicular to normal of both planes. i.e. š‘ āƒ— is perpendicular to normal of both planes. The vector equation of a line passing through a point with position vector š‘Ž āƒ— and parallel to a vector š‘ āƒ— is š’“ āƒ— = š’‚ āƒ— + šœ†š’ƒ āƒ— Given, the line passes through (1, 2, 3) So, š‘Ž āƒ— = 1š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ Given, line is parallel to both planes āˆ“ Line is perpendicular to normal of both planes. i.e. š‘ āƒ— is perpendicular to normal of both planes. We know that š‘Ž āƒ— Ɨ š‘ āƒ— is perpendicular to both š‘Ž āƒ— & š‘ āƒ— So, š‘ āƒ— is cross product of normal of planes š‘Ÿ āƒ— . (š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 2 š‘˜ Ģ‚) = 5 and š‘Ÿ āƒ— . (3š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚) = 6 Required normal = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@1&āˆ’1&2@3&1&1)| = š‘– Ģ‚ (ā€“1(1) ā€“ 1(2)) ā€“ š‘— Ģ‚ (1(1) ā€“ 3(2)) + š‘˜ Ģ‚(1(1) ā€“ 3(ā€“1)) = š‘– Ģ‚ (ā€“1 ā€“ 2) ā€“ š‘— Ģ‚ (1 ā€“ 6) + š‘˜ Ģ‚(1 + 3) = ā€“3š‘– Ģ‚ + 5š‘— Ģ‚ + 4š‘˜ Ģ‚ Thus, š‘ āƒ— = āˆ’3š‘– Ģ‚ + 5š‘— Ģ‚ + 4š‘˜ Ģ‚ Now, Putting value of š‘Ž āƒ— & š‘ āƒ— in formula š‘Ÿ āƒ— = š‘Ž āƒ— + šœ†š‘ āƒ— = (š’Š Ģ‚ + 2š’‹ Ģ‚ + 3š’Œ Ģ‚) + šœ† (āˆ’3š’Š Ģ‚ + 5š’‹ Ģ‚ + 4š’Œ Ģ‚) Now, Putting value of š‘Ž āƒ— & š‘ āƒ— in formula š‘Ÿ āƒ— = š‘Ž āƒ— + šœ†š‘ āƒ— = (š’Š Ģ‚ + 2š’‹ Ģ‚ + 3š’Œ Ģ‚) + šœ† (āˆ’3š’Š Ģ‚ + 5š’‹ Ģ‚ + 4š’Œ Ģ‚) Misc 19 (Method 2) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes š‘Ÿ āƒ— . (š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 2 š‘˜ Ģ‚) = 5 and š‘Ÿ āƒ— . (3š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚) = 6 . The vector equation of a line passing through a point with position vector š‘Ž āƒ— and parallel to a vector š‘ āƒ— is š’“ āƒ— = š’‚ āƒ— + šœ†š’ƒ āƒ— Given, the line passes through (1, 2, 3) So, š‘Ž āƒ— = 1š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ Let š‘ āƒ— = š‘_1 š‘– Ģ‚ + š‘_2 š‘— Ģ‚ + š‘_3 š‘˜ Ģ‚ A line parallel to a plane is perpendicular to the normal of the plane. And two lines š‘ āƒ— and š‘ž āƒ— are perpendicular if š‘ āƒ—.š‘ž āƒ— = 0 Given, the line is parallel to planes š’“ āƒ—.(š’Š Ģ‚ āˆ’ š’‹ Ģ‚ + 2š’Œ Ģ‚) = 5 Comparing with š‘Ÿ āƒ—. (š‘›1) āƒ— = d1, (š‘›1) āƒ— = 1š‘– Ģ‚ āˆ’ 1š‘— Ģ‚ + 2š‘˜ Ģ‚ Since š‘ āƒ— is āŠ„ to (š‘›1) āƒ—, š‘ āƒ—.(š‘›1) āƒ— = 0 (š‘1š‘– Ģ‚ + š‘2 š‘— Ģ‚ + š‘3 š‘˜ Ģ‚).(1š‘– Ģ‚ āˆ’ 1š‘— Ģ‚ + 2š‘˜ Ģ‚) = 0 (š‘"1"Ɨ 1) + (š‘"2"Ɨ āˆ’1) + (š‘3 Ɨ 2) = 0 š’ƒ1 āˆ’ š’ƒ2 + 2š’ƒ3 = 0 š’“ āƒ—.(3š’Š Ģ‚ + š’‹ Ģ‚ + š’Œ Ģ‚) = 6 Comparing with š‘Ÿ āƒ—. (š‘›2) āƒ— = d2, (š‘›2) āƒ— = 3š‘– Ģ‚ + 1š‘— Ģ‚ + 1š‘˜ Ģ‚ Since š‘ āƒ— is āŠ„ to (š‘›2) āƒ—, š‘ āƒ—.(š‘›2) āƒ— = 0 (š‘1 š‘– Ģ‚ + š‘2 š‘— Ģ‚ + š‘3 š‘˜ Ģ‚).(3š‘– Ģ‚ + 1š‘— Ģ‚ + 1š‘˜ Ģ‚) = 0 (š‘1 Ɨ 3) + (š‘2 Ɨ 1) + (š‘3 Ɨ 1) = 0 3š’ƒ1 + š’ƒ2 + š’ƒ3 = 0 So, our equations are š‘1 āˆ’ š‘2 + 2š‘3 = 0 3š‘1 + š‘2 + š‘3 = 0 Thus, š‘ āƒ— = š‘_1 š‘– Ģ‚ + š‘_2 š‘— Ģ‚ + š‘_3 š‘˜ Ģ‚ = āˆ’3kš‘– Ģ‚ + 5kš‘— Ģ‚ + 4kš‘˜ Ģ‚ Now, Putting value of š‘Ž āƒ— & š‘ āƒ— in formula š‘Ÿ āƒ— = š‘Ž āƒ— + šœ†š‘ āƒ— āˆ“ š‘Ÿ āƒ— = (1š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚) + šœ† (āˆ’3kš‘– Ģ‚ + 5kš‘— Ģ‚ + 4kš‘˜ Ģ‚) = (š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚) + šœ†k (āˆ’3š‘– Ģ‚ + 5š‘— Ģ‚ + 4š‘˜ Ģ‚) = (š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚) + šœ† (āˆ’3š‘– Ģ‚ + 5š‘— Ģ‚ + 4š‘˜ Ģ‚) Therefore, the equation of the line is (š’Š Ģ‚ + 2š’‹ Ģ‚ + 3š’Œ Ģ‚) + šœ† (āˆ’3š’Š Ģ‚ + 5š’‹ Ģ‚ + 4š’Œ Ģ‚).

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.