Question
Let P be the plane 3 x + 2 y + 3 z = 16 and let S = α i ^ + β j ^ + γ k ^ :   α 2 + β 2 + γ 2 = 1 and the distance of α ,   β ,   γ from the plane P is 7 2 . Let u → ,   v → and w → be three distinct vectors in S such that u → - v → = v → - w → = w → - u → . Let V be the volume of the parallelepiped determined by vectors u → ,   v → and w → . Then the value of 80 3 V is
Step-by-step solution
Given, Equation of plane, P : 3 x + 2 y + 3 z = 16 And S = α i ^ + β j ^ + γ k ^   :   α 2 + β 2 + γ 2 = 1 which is equation of sphere, Also distance, d α ,   β ,   γ from P = 7 2 And relation between distinct vector is given by, u → - v → = v → - w → = w → - u → Now V : volume of parallelepiped by vectors u → ,   v → ,   w → So, using formula of distance d α ,   β ,   γ from P = 7 2 we get, ⇒ 3 α + 2 β + 3 γ - 16 3 + 4 + 9 = 7 2 ⇒ 3 α + 2 β + 3 γ - 16 4 = 7 2 ⇒ 3 α + 2 β + 3 γ - 16 = 14   . . . . i Also, α 2 + β 2 + γ 2 = 1   . . . . . . . ii Now we know that, Volume of parallelepiped by vector, u → ,   v → ,   w → is given by, V = u →   v →   w → = u → · v × w   . . . . . . iii u → = v → = w → = 1 (As they lie on sphere of unit radius) . . . . . . iv u → - v → = v → - w → = w → - u → (Given) Now squaring, we get, ⇒ u → - v → 2 = v → - w → 2 = w → - u → 2 ⇒ u 2 + v 2 - 2 u → · v → A = v 2 + w 2 - 2 v → · w → B = w 2 + u 2 - 2 w → · u → ⏟ C Now from A and B we get, ⇒ u 2 + v 2 - 2 u → · v → = v 2 + w 2 - 2 v → · w → ⇒ u 2 - w 2 = 2 u → · v → - 2 v → · w → ∵   u → = w → = 1   Given ⇒ u → · v → = v → · w → Hence, by using B and C also, we will get u → · v → = v → · w → = w → · u → = m   say   . . . . . . . v ⇒ u → ,   v → ,   w → are the vectors of an equilateral triangle (say ∆   A B C ) d O ,   P = 16 3 + 4 + 9 = 16 4 = 4 units O A → = u → ,   O B → = v → ,   O C → = w → O A → = O B → = O C → = 1 (Given) In an equilateral triangle, circumcentre, orthrocentre and centroid coincide. Let D be the circumcentre of &