Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
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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 + šš 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 + šš 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&ā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 ā = ā3š Ģ + 5š Ģ + 4š Ģ Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā = (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ) Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā = (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š 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 + šš 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. (š1) ā = d1, (š1) ā = 1š Ģ ā 1š Ģ + 2š Ģ Since š ā is ā„ to (š1) ā, š ā.(š1) ā = 0 (š1š Ģ + š2 š Ģ + š3 š Ģ).(1š Ģ ā 1š Ģ + 2š Ģ) = 0 (š"1"Ć 1) + (š"2"Ć ā1) + (š3 Ć 2) = 0 š1 ā š2 + 2š3 = 0 š ā.(3š Ģ + š Ģ + š Ģ) = 6 Comparing with š ā. (š2) ā = d2, (š2) ā = 3š Ģ + 1š Ģ + 1š Ģ Since š ā is ā„ to (š2) ā, š ā.(š2) ā = 0 (š1 š Ģ + š2 š Ģ + š3 š Ģ).(3š Ģ + 1š Ģ + 1š Ģ) = 0 (š1 Ć 3) + (š2 Ć 1) + (š3 Ć 1) = 0 3š1 + š2 + š3 = 0 So, our equations are š1 ā š2 + 2š3 = 0 3š1 + š2 + š3 = 0 Thus, š ā = š_1 š Ģ + š_2 š Ģ + š_3 š Ģ = ā3kš Ģ + 5kš Ģ + 4kš Ģ Now, Putting value of š ā & š ā in formula š ā = š ā + šš ā ā“ š ā = (1š Ģ + 2š Ģ + 3š Ģ) + š (ā3kš Ģ + 5kš Ģ + 4kš Ģ) = (š Ģ + 2š Ģ + 3š Ģ) + šk (ā3š Ģ + 5š Ģ + 4š Ģ) = (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ) Therefore, the equation of the line is (š Ģ + 2š Ģ + 3š Ģ) + š (ā3š Ģ + 5š Ģ + 4š Ģ).