Exams Joint Entrance Examination Mathematics Vector Algebra Question 239 Question 239 of 379 Medium NUMERICAL
Let p ⃗ = 2 i ^ + 3 j ^ + k ^ \vec p=2\hat i+3\hat j+\hat k p = 2 i ^ + 3 j ^ + k ^ , q ⃗ = i ^ + 2 j ^ + k ^ \vec q=\hat i+2\hat j+\hat k q = i ^ + 2 j ^ + k ^ . If r ⃗ = α i ^ + β j ^ + γ k ^ \vec r=\alpha\hat i+\beta\hat j+\gamma\hat k r = α i ^ + β j ^ + γ k ^ is perpendicular to both p ⃗ + q ⃗ \vec p+\vec q p + q and p ⃗ − q ⃗ \vec p-\vec q p − q , and ∣ r ⃗ ∣ = 3 |\vec r|=\sqrt{3} ∣ r ∣ = 3 , then ∣ α ∣ + ∣ β ∣ + ∣ γ ∣ |\alpha|+|\beta|+|\gamma| ∣ α ∣ + ∣ β ∣ + ∣ γ ∣ is \_\_\_\_\_\_\_.
Answer: 3
Solutions · 1 Post your solution Official solution
p ⃗ + q ⃗ = ( 3 , 5 , 2 ) \vec p+\vec q=(3,5,2) p + q = ( 3 , 5 , 2 ) , p ⃗ − q ⃗ = ( 1 , 1 , 0 ) \vec p-\vec q=(1,1,0) p − q = ( 1 , 1 , 0 ) .
\[(\vec p+\vec q)\times(\vec p-\vec q) = \begin{vmatrix}\hat i&\hat j&\hat k\\3&5&2\\1&1&0\end{vmatrix} = (-2,2,-2)\]
r ⃗ = ± 3 ⋅ ( − 2 , 2 , − 2 ) 12 = ± ( − 1 , 1 , − 1 ) \vec r = \pm\sqrt{3}\cdot\dfrac{(-2,2,-2)}{\sqrt{12}} = \pm(-1,1,-1) r = ± 3 ⋅ 12 ( − 2 , 2 , − 2 ) = ± ( − 1 , 1 , − 1 ) , giving ∣ α ∣ = ∣ β ∣ = ∣ γ ∣ = 1 |\alpha|=|\beta|=|\gamma|=1 ∣ α ∣ = ∣ β ∣ = ∣ γ ∣ = 1 .
\[|\alpha|+|\beta|+|\gamma| = 3\] Answer the question to unlock