Let P be the plane passing through the intersection of the planes
r→.(i+3k−k)=5 and r→ .(2i−j+k)=3,
and the point (2, 1, –2). Let the position vectors of the points X and Y be
i−2j+4k and 5i−j+2k
respectively. Then the points
The correct option is(C): X and Y are on the opposite sides of P.
Let the equation of required plane
\(\pi:(x+3y-z-5)+λ(2x-y+z-3)=0\)
\(∵(2,1,-2)\,\text{lies on it so,} 2+λ(-2)=0\)
⇒λ=1
Hence,
\(\pi:3x+2y-8=0\)
\(∵\pi{x}=-9,\pi{y},\pi_{x+y}=4\)
\(\pi_{x+y}=-22\,and\,\pi_{y-x}=6\)
Clearly, X and Y are on opposite sides of plane π.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely: