Check sibling questions

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

This video is only available for Teachoo black users

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

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‚).

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.