Let P be the foot of the perpendicular from the point Q ( 10 , − 3 , − 1 ) on the line 7 x − 3 = − 1 y − 2 = − 2 z + 1 . Then the area of the right angled triangle PQR , where R is the point ( 3

Question

Let P be the foot of the perpendicular from the point Q ( 10 , − 3 , − 1 ) on the line 7 x − 3 ​ = − 1 y − 2 ​ = − 2 z + 1 ​ . Then the area of the right angled triangle PQR , where R is the point ( 3 , − 2 , 1 ) , is

TestHub · Accepted Answer

​ 7 x − 3 ​ = − 1 y − 2 ​ = − 2 z + 1 ​ = λ ⇒ 7 λ + 3 , − λ + 2 , − 2 λ − 1 dr’s of QP ⇒ 7 λ − 7 , − λ + 5 , − 2 λ ​ Now ​ ( 7 λ − 7 ) ⋅ 7 − ( − λ + 5 ) + ( 2 λ ) ⋅ 2 = 0 54 λ − 54 = 0 ⇒ λ = 1 ∴ P = ( 10 , 1 , − 3 ) PQ ​ = − 4 j ^ ​ + 2 k ^ PR = − 7 i ^ − 3 j ^ ​ + 4 k ^ Area = 2 1 ​ ​ i 0 − 7 ​ j − 4 − 3 ​ k 2 4 ​ ​ = 3 30 ​ ​

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