Exams Joint Entrance Examination Mathematics 3D Geometry Question 214 Question 214 of 521 Medium NUMERICAL
If λ 1 < λ 2 \lambda_1<\lambda_2 λ 1 < λ 2 are two values of λ \lambda λ such that the angle between the planes P 1 : r ⃗ ⋅ ( 3 i ^ − 5 j ^ + k ^ ) = 7 P_1:\ \vec r\cdot(3\hat i-5\hat j+\hat k)=7 P 1 : r ⋅ ( 3 i ^ − 5 j ^ + k ^ ) = 7 and P 2 : r ⃗ ⋅ ( λ i ^ + j ^ − 3 k ^ ) = 9 P_2:\ \vec r\cdot(\lambda\hat i+\hat j-3\hat k)=9 P 2 : r ⋅ ( λ i ^ + j ^ − 3 k ^ ) = 9 is sin − 1 ( 2 6 5 ) \sin^{-1}\left(\dfrac{2\sqrt{6}}{5}\right) sin − 1 ( 5 2 6 ) , then the square of the length of the perpendicular from the point ( 38 λ 1 , 10 λ 2 , 2 ) (38\lambda_1,\ 10\lambda_2,\ 2) ( 38 λ 1 , 10 λ 2 , 2 ) to the plane P 1 P_1 P 1 is:
Answer: 315
Solutions · 1 Post your solution Official solution
sin θ = 2 6 5 ⇒ cos θ = 1 5 \sin\theta=\dfrac{2\sqrt{6}}{5}\Rightarrow\cos\theta=\dfrac15 sin θ = 5 2 6 ⇒ cos θ = 5 1 .
\[\cos\theta=\frac{3\lambda-5-3}{\sqrt{35}\sqrt{\lambda^2+10}}=\frac{3\lambda-8}{\sqrt{35}\sqrt{\lambda^2+10}}=\frac15.\]
Squaring: 25 ( 3 λ − 8 ) 2 = 35 ( λ 2 + 10 ) ⟹ 190 λ 2 − 1200 λ + 1250 = 0 ⟹ 19 λ 2 − 120 λ + 125 = 0. 25(3\lambda-8)^2=35(\lambda^2+10) \implies 190\lambda^2-1200\lambda+1250=0 \implies 19\lambda^2-120\lambda+125=0. 25 ( 3 λ − 8 ) 2 = 35 ( λ 2 + 10 ) ⟹ 190 λ 2 − 1200 λ + 1250 = 0 ⟹ 19 λ 2 − 120 λ + 125 = 0.
\[\lambda=\frac{120\pm70}{38}=5\ \text{or}\ \frac{25}{19}.\]
So λ 1 = 25 19 , λ 2 = 5 \lambda_1=\dfrac{25}{19},\ \lambda_2=5 λ 1 = 19 25 , λ 2 = 5 .
Point: ( 38 λ 1 , 10 λ 2 , 2 ) = ( 50 , 50 , 2 ) (38\lambda_1,10\lambda_2,2)=(50,50,2) ( 38 λ 1 , 10 λ 2 , 2 ) = ( 50 , 50 , 2 ) .
\[d=\frac{|3(50)-5(50)+2-7|}{\sqrt{35}}=\frac{105}{\sqrt{35}}=3\sqrt{35}.\]
\[d^2=9\times35=315.\] Answer the question to unlock