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Showing posts with the label vector form of lines

Consider the lines L₁ : x − 1 = y − 2 = z and L₂ : x − 2 = y = z − 1. Let the feet of the perpendiculars from the point P(5, 1, −3) on the lines L₁ and L₂ be Q and R respectively. If the area of the triangle PQR is A, then 4A² is equal to:

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  Question: Consider the lines L 1 : x − 1 = y − 2 = z L_{1} : x - 1 = y - 2 = z and L 2 : x − 2 = y = z − 1 L_{2} : x - 2 = y = z - 1 Let the feet of the perpendiculars from the point P ( 5 ,   1 ,   − 3 ) P(5,\,1,\,-3)  on the lines L 1 L_{1} ​ and L 2 L_{2} ​ be Q Q  and R R  respectively. If the area of the triangle P Q R PQR  is A A , then find 4 A 2 4A^{2} . Question Image Short Solution To find 4 A 2 4A^{2} , follow these steps: Write lines in parametric form L 1 :    x = 1 + t ,    y = 2 + t ,    z = t L 2 :    x = 2 + s ,    y = s ,    z = 1 + s Find the foot of the perpendicular from P P  to each line Use the formula for foot of perpendicular on a line: Q = A + ( P − A ) ⋅ d ∣ d ∣ 2 d Q = A + \dfrac{(P-A)\cdot d}{|d|^{2}} d where A A  is a point on the line and d d  its direction vector. Get coordinates of Q Q  and R R  by solving the dot product condition. Find the area of triangle P Q R P...