Exams Joint Entrance Examination Mathematics Vector Algebra Question 148 Question 148 of 379 Medium NUMERICAL
Let a ⃗ = i ^ + j ^ + k ^ \vec a=\hat i+\hat j+\hat k a = i ^ + j ^ + k ^ , b ⃗ = − i ^ − 8 j ^ + 2 k ^ \vec b=-\hat i-8\hat j+2\hat k b = − i ^ − 8 j ^ + 2 k ^ , and c ⃗ = 4 i ^ + c 2 j ^ + c 3 k ^ \vec c=4\hat i+c_2\hat j+c_3\hat k c = 4 i ^ + c 2 j ^ + c 3 k ^ be vectors such that b ⃗ × a ⃗ = c ⃗ × a ⃗ \vec b\times\vec a=\vec c\times\vec a b × a = c × a . If the angle between c ⃗ \vec c c and 3 i ^ + 4 j ^ + k ^ 3\hat i+4\hat j+\hat k 3 i ^ + 4 j ^ + k ^ is θ \theta θ , the greatest integer ≤ tan 2 θ \le\tan^2\theta ≤ tan 2 θ is:
Answer: 38
Solutions · 1 Post your solution Official solution
( b ⃗ − c ⃗ ) × a ⃗ = 0 ⃗ ⇒ b ⃗ − c ⃗ = λ a ⃗ (\vec b-\vec c)\times\vec a=\vec 0 \Rightarrow \vec b-\vec c = \lambda\vec a ( b − c ) × a = 0 ⇒ b − c = λ a :
\[-\hat i-8\hat j+2\hat k = 4\hat i+c_2\hat j+c_3\hat k+\lambda(\hat i+\hat j+\hat k)\]
λ + 4 = − 1 ⇒ λ = − 5 \lambda+4=-1\Rightarrow\lambda=-5 λ + 4 = − 1 ⇒ λ = − 5 ; λ + c 2 = − 8 ⇒ c 2 = − 3 \lambda+c_2=-8\Rightarrow c_2=-3 λ + c 2 = − 8 ⇒ c 2 = − 3 ; λ + c 3 = 2 ⇒ c 3 = 7 \lambda+c_3=2\Rightarrow c_3=7 λ + c 3 = 2 ⇒ c 3 = 7 . So c ⃗ = 4 i ^ − 3 j ^ + 7 k ^ \vec c=4\hat i-3\hat j+7\hat k c = 4 i ^ − 3 j ^ + 7 k ^ .
\[\cos\theta = \frac{12-12+7}{\sqrt{26}\sqrt{74}} = \frac{7}{2\sqrt{481}}\]
\[\tan^2\theta = \sec^2\theta-1 = \frac{4\cdot481}{49}-1 = \frac{1924-49}{49}=\frac{1875}{49}\approx38.27\]
\[\lfloor\tan^2\theta\rfloor = 38\] Answer the question to unlock