Exams Joint Entrance Examination Mathematics 3D Geometry Question 132 Question 132 of 521 Medium NUMERICAL
Let Q Q Q and R R R be the feet of perpendiculars from the point P ( a , a , a ) P(a,a,a) P ( a , a , a ) on the lines x = y , z = 1 x=y,z=1 x = y , z = 1 and x = − y , z = − 1 x=-y,z=-1 x = − y , z = − 1 respectively. If \angleQPR \angleQPR \angleQPR is a right angle, then 12 a 2 12{a}^{2} 12 a 2 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 12
Solutions · 1 Post your solution Official solution
x 1 = y 1 = z − 1 0 = r → Q ( r , r , 1 ) x 1 = y − 1 = z + 1 0 = k → R ( k , − k , − 1 ) P Q ⃗ = ( a − r ) i ^ + ( a − r ) j ^ + ( a − 1 ) k ^ a = r + a − r = 0 2 a = 2 r → a = r P R ⃗ = ( a − k ) i + ( a + k ) j ^ + ( a + 1 ) k ^ a − k − a − k = 0 ⇒ k = 0 A s , P Q ⊥ P R ( a − r ) ( a − k ) + ( a − r ) ( a + k ) + ( a − 1 ) ( a + 1 ) = 0 a = 1 − 1 12 a 2 = 12 \begin{gathered} \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r\to Q(r,r,1) \\ \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k\to R(k,-k,-1) \\ \vec{PQ}=(a-r)\hat{i}+(a-r)\hat{j}+(a-1)\hat{k} \\ a=r+a-r=0 \\ 2a=2r\to a=r \\ \vec{PR}=(a-k)i+(a+k)\hat{j}+(a+1)\hat{k} \\ a-k-a-k=0\Rightarrow k=0 \\ As,PQ\perp PR \\ (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\ a=1-1 \\ 12{a}^{2}=12 \end{gathered} 1 x = 1 y = 0 z − 1 = r → Q ( r , r , 1 ) 1 x = − 1 y = 0 z + 1 = k → R ( k , − k , − 1 ) P Q = ( a − r ) i ^ + ( a − r ) j ^ + ( a − 1 ) k ^ a = r + a − r = 0 2 a = 2 r → a = r P R = ( a − k ) i + ( a + k ) j ^ + ( a + 1 ) k ^ a − k − a − k = 0 ⇒ k = 0 A s , P Q ⊥ P R ( a − r ) ( a − k ) + ( a − r ) ( a + k ) + ( a − 1 ) ( a + 1 ) = 0 a = 1 − 1 12 a 2 = 12 Answer the question to unlock