Miscellaneous
Miscellaneous
Last updated at December 16, 2024 by Teachoo
Transcript
Question 7 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.The equation of a line passing through two points with position vectors š ā & š ā is š ā = š ā + š(š ā ā š ā) Given, the line passes through (š ā ā š ā) = (3š Ģ + 4š Ģ + 1š Ģ) ā (5š Ģ + 1š Ģ + 6š Ģ) = (3 ā5)š Ģ + (4 ā 1)š Ģ + (1 ā 6)š Ģ A (5, 1, 6) š ā = 5š Ģ + 1š Ģ + 6š Ģ B(3, 4, 1) š ā = 3š Ģ + 4š Ģ + 1š Ģ = ā2š Ģ + 3š Ģ ā 5š Ģ ā“ š ā = (5š Ģ + š Ģ + 6š Ģ) + š (ā2š Ģ + 3š Ģ ā 5š Ģ) Let the coordinates of the point where the line crosses the ZX plane be (x, 0, z) So, š ā = xš Ģ + 0š Ģ + zš Ģ Since point lies in line, it will satisfy its equation, Putting (2) in (1) xš Ģ + 0š Ģ + zš Ģ = 5š Ģ + š Ģ + 6š Ģ ā2šš Ģ + 3šš Ģ ā 5šš Ģ xš Ģ + 0š Ģ + zš Ģ = (5 ā2š)š Ģ + (1 + 3š)š Ģ + (6 ā 5š)š Ģ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 3š 3š = ā1 ā“ š = (āš)/š Now, x = 5 ā 2š = 5 ā 2 Ć (ā1)/3 = 5 + 2/3 = 17/13 z = 6 ā 5š = 6 ā 5 Ć (ā1)/3 = 6 + 5/3 = 23/3 Therefore, the coordinate of the required point are (šš/š,š,šš/š) Question 7 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.The equation of a line passing through two points A(š„_1, š¦_1, š§_1) and B(š„_2, š¦_2, š§_2) is (š ā š_š)/(š_š ā š_š ) = (š ā š_š)/(š_š ā š_š ) = (š ā š_š)/(š_š ā š_š ) Given the line passes through the points A (5, 1, 6) ā“ š„_1= 5, š¦_1= 1, š§_1= 6 B(3, 4, 1) ā“ š„_2= 3, š¦_2= 4, š§_2= 1 So, the equation of line is (š„ ā 5)/(3 ā 5) = (š¦ ā 1)/(4 ā 1) = (š§ ā 6)/(1 ā 6) (š ā š)/(āš) = (š ā š)/š = (š ā š)/(āš) = k So, Since the line crosses the ZX plane at (x, 0, z), y = 0 3k + 1 = 0 3k = ā1 ā“ k = (āš)/š So, x = ā2k + 5 = ā2 Ć (ā1)/3 + 5 = 2/3 + 5 = 17/3 y = 0 & z = ā5k + 6 = ā5 Ć (ā1)/3 + 6 = 5/3 + 6 = 23/3 therefore, the coordinate of the required point are (šš/š,š,šš/š)