Exams Joint Entrance Examination Mathematics 3D Geometry Question 293 Question 293 of 521 Medium NUMERICAL
If the shortest distance between the lines r ⃗ 1 = ( α i ^ + 2 j ^ + 2 k ^ ) + λ ( i ^ − 2 j ^ + 2 k ^ ) \vec r_1=(\alpha\hat i+2\hat j+2\hat k)+\lambda(\hat i-2\hat j+2\hat k) r 1 = ( α i ^ + 2 j ^ + 2 k ^ ) + λ ( i ^ − 2 j ^ + 2 k ^ ) , λ ∈ R \lambda\in\mathbb R λ ∈ R , α > 0 \alpha>0 α > 0 , and r ⃗ 2 = ( − 4 i ^ − k ^ ) + μ ( 3 i ^ − 2 j ^ − 2 k ^ ) \vec r_2=(-4\hat i-\hat k)+\mu(3\hat i-2\hat j-2\hat k) r 2 = ( − 4 i ^ − k ^ ) + μ ( 3 i ^ − 2 j ^ − 2 k ^ ) , μ ∈ R \mu\in\mathbb R μ ∈ R , is 9 9 9 , then α \alpha α is equal to:
Answer: 6
Solutions · 1 Post your solution Official solution
a ⃗ − c ⃗ = ( α + 4 , 2 , 3 ) \vec a-\vec c=(\alpha+4,2,3) a − c = ( α + 4 , 2 , 3 ) . b ⃗ × d ⃗ ∝ ( 2 , 2 , 1 ) \vec b\times\vec d\ \propto\ (2,2,1) b × d ∝ ( 2 , 2 , 1 ) , with ∣ b ⃗ × d ⃗ ∣ = 3 |\vec b\times\vec d|=3 ∣ b × d ∣ = 3 .
\[\frac{(\alpha+4,2,3)\cdot(2,2,1)}{3}=9 \implies 2\alpha+8+4+3=27 \implies 2\alpha+15=27 \implies \alpha=6.\] Answer the question to unlock