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Find the vector equation of the plane that passes through the point (1, 0, 0) and contains the line r
^{
→
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= βi Μ j Μ

Last updated at Oct. 27, 2020 by Teachoo

Transcript

Equation of plane passing through point A whose position vector is π β & perpendicular to π β is (π β β π β) . π β = 0 Given Plane passes through (1, 0, 0) So π β = 1π Μ + 0π Μ β 0π Μ π β = π Μ Thus, equation of plane is (π β β π β) . π β = 0 (π β β π Μ) . π β = 0 Now, The plane contains the line π β= ππ Μ So, Line is perpendicular to normal of plane π Μ . π β = 0 And, Lines point (0, 0, 0) lies on plane as well Putting π β = 0 β in equation of plane (0 β β π Μ) . π β = 0 β π Μ . π β = 0 π Μ . π β = 0 Since π β perpendicular to both π Μ and π Μ Thus, π β = π Μ So, our equation of plane becomes (π β β π Μ) . π β = 0 (π β β π Μ) . π Μ = 0 π β . π Μ β π Μ . π Μ = 0 π β . π Μ = 0

CBSE Class 12 Sample Paper for 2021 Boards

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Class 12

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.