Exams Joint Entrance Examination Mathematics Vector Algebra Question 234 Question 234 of 379 Medium MCQ
Let O O O be the origin, O P ⃗ = x i ^ + y j ^ − k ^ \vec{OP}=x\hat i+y\hat j-\hat k O P = x i ^ + y j ^ − k ^ , O Q ⃗ = − i ^ + 2 j ^ + 3 x k ^ \vec{OQ}=-\hat i+2\hat j+3x\hat k O Q = − i ^ + 2 j ^ + 3 x k ^ ( x , y ∈ R , x > 0 x,y\in\mathbb R,x>0 x , y ∈ R , x > 0 ), with ∣ P Q ∣ = 20 |PQ|=\sqrt{20} ∣ P Q ∣ = 20 and O P ⃗ ⊥ O Q ⃗ \vec{OP}\perp\vec{OQ} O P ⊥ O Q . If O R ⃗ = 3 i ^ + z j ^ − 7 k ^ \vec{OR}=3\hat i+z\hat j-7\hat k O R = 3 i ^ + z j ^ − 7 k ^ is coplanar with O P ⃗ , O Q ⃗ \vec{OP},\vec{OQ} O P , O Q , then x 2 + y 2 + z 2 x^2+y^2+z^2 x 2 + y 2 + z 2 is:
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P Q ⃗ = ( − 1 − x , 2 − y , 3 x + 1 ) \vec{PQ}=(-1-x,\ 2-y,\ 3x+1) P Q = ( − 1 − x , 2 − y , 3 x + 1 ) , ∣ P Q ∣ 2 = 20 |PQ|^2=20 ∣ P Q ∣ 2 = 20 :
( x + 1 ) 2 + ( y − 2 ) 2 + ( 3 x + 1 ) 2 = 20 (x+1)^2+(y-2)^2+(3x+1)^2=20 ( x + 1 ) 2 + ( y − 2 ) 2 + ( 3 x + 1 ) 2 = 20
O P ⃗ ⋅ O Q ⃗ = 0 \vec{OP}\cdot\vec{OQ}=0 O P ⋅ O Q = 0 : − x + 2 y − 3 x = 0 ⇒ y = 2 x -x+2y-3x=0 \Rightarrow y=2x − x + 2 y − 3 x = 0 ⇒ y = 2 x . Substituting:
\[(x+1)^2+(2x-2)^2+(3x+1)^2=20 \implies 14x^2+6=20 \implies x^2=1 \implies x=1\ (x>0)\]
So y = 2 y=2 y = 2 . Coplanarity: [ O P ⃗ O Q ⃗ O R ⃗ ] = 0 [\vec{OP}\ \vec{OQ}\ \vec{OR}]=0 [ O P O Q O R ] = 0 :
\[\begin{vmatrix}1&2&-1\\-1&2&3\\3&z&-7\end{vmatrix}=0 \implies z=-2\]
x 2 + y 2 + z 2 = 1 + 4 + 4 = 9 x^2+y^2+z^2 = 1+4+4=9 x 2 + y 2 + z 2 = 1 + 4 + 4 = 9 Answer the question to unlock