Equation of line under planes condition
Equation of line under planes condition
Last updated at April 16, 2024 by Teachoo
Question 15 (Method 1) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes š ā . (š Ģ ā š Ģ + 2 š Ģ) = 5 and š ā . (3š Ģ + š Ģ + š Ģ) = 6 . The vector equation of a line passing through a point with position vector š ā and parallel to a vector š ā is š ā = š ā + šš ā Given, the line passes through (1, 2, 3) So, š ā = 1š Ģ + 2š Ģ + 3š Ģ Given, line is parallel to both planes ā“ Line is perpendicular to normal of both planes. i.e. š ā is perpendicular to normal of both planes. The vector equation of a line passing through a point with position vector š ā and parallel to a vector š ā is š ā = š ā + šš ā Given, the line passes through (1, 2, 3) So, š ā = 1š Ģ + 2š Ģ + 3š Ģ Given, line is parallel to both planes ā“ Line is perpendicular to normal of both planes. i.e. š ā is perpendicular to normal of both planes. We know that š ā Ć š ā is perpendicular to both š ā & š ā So, š ā is cross product of normal of planes š ā . (š Ģ ā š Ģ + 2 š Ģ) = 5 and š ā . (3š Ģ + š Ģ + š Ģ) = 6 Required normal = |ā 8(š Ģ&š Ģ&š Ģ@1&ā1&2@3&1&1)| = š Ģ (ā1(1) ā 1(2)) ā š Ģ (1(1) ā 3(2)) + š Ģ(1(1) ā 3(ā1)) = š Ģ (ā1 ā 2) ā š Ģ (1 ā 6) + š Ģ(1 + 3) = ā3š Ģ + 5š Ģ + 4š Ģ Thus, š ā = ā3š Ģ + 5š Ģ + 4š Ģ Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā = (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ) Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā = (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ) Question 15 (Method 2) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes š ā . (š Ģ ā š Ģ + 2 š Ģ) = 5 and š ā . (3š Ģ + š Ģ + š Ģ) = 6 . The vector equation of a line passing through a point with position vector š ā and parallel to a vector š ā is š ā = š ā + šš ā Given, the line passes through (1, 2, 3) So, š ā = 1š Ģ + 2š Ģ + 3š Ģ Let š ā = 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 ā and š ā are perpendicular if š ā.š ā = 0 Given, the line is parallel to planes š ā.(š Ģ ā š Ģ + 2š Ģ) = 5 Comparing with š ā. (š1) ā = d1, (š1) ā = 1š Ģ ā 1š Ģ + 2š Ģ Since š ā is ā„ to (š1) ā, š ā.(š1) ā = 0 (š1š Ģ + š2 š Ģ + š3 š Ģ).(1š Ģ ā 1š Ģ + 2š Ģ) = 0 (š"1"Ć 1) + (š"2"Ć ā1) + (š3 Ć 2) = 0 š1 ā š2 + 2š3 = 0 š ā.(3š Ģ + š Ģ + š Ģ) = 6 Comparing with š ā. (š2) ā = d2, (š2) ā = 3š Ģ + 1š Ģ + 1š Ģ Since š ā is ā„ to (š2) ā, š ā.(š2) ā = 0 (š1 š Ģ + š2 š Ģ + š3 š Ģ).(3š Ģ + 1š Ģ + 1š Ģ) = 0 (š1 Ć 3) + (š2 Ć 1) + (š3 Ć 1) = 0 3š1 + š2 + š3 = 0 So, our equations are š1 ā š2 + 2š3 = 0 3š1 + š2 + š3 = 0 Thus, š ā = š_1 š Ģ + š_2 š Ģ + š_3 š Ģ = ā3kš Ģ + 5kš Ģ + 4kš Ģ Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā ā“ š ā = (1š Ģ + 2š Ģ + 3š Ģ) + š (ā3kš Ģ + 5kš Ģ + 4kš Ģ) = (š Ģ + 2š Ģ + 3š Ģ) + šk (ā3š Ģ + 5š Ģ + 4š Ģ) = (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ) Therefore, the equation of the line is (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ).