JEE Advanced2018MathematicsVector AlgebraActual
Consider the cube in the first octant with sides O P , O Q and OR of length 1 , along the x-axis, y-axis and z-axis, respectively, where O ( 0 , 0 , 0 ) is the origin. Let S 1 2 , 1 2 , 1 2 be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal O T . If p → = S P → , q → = S Q → , r → = S R → and t → = S T → ,
Correct answer
0
Step-by-step solution
p → = S P → = 1 2 , - 1 2 , - 1 2 = 1 2 i ^ - j ^ - k ^ q → = S Q → = - 1 2 , 1 2 , - 1 2 = 1 2 - i ^ + j ^ - k ^ r → = S R → = - 1 2 , - 1 2 , 1 2 = 1 2 - i ^ - j ^ + k ^ t → = S T → = 1 2 , 1 2 , 1 2 = 1 2 i ^ + j ^ + k ^ p → × q → × r → × t → = 1 4 i ^ j ^ k ^ 1 - 1 - 1 - 1 1 - 1 × 1 4 i ^ j ^ k ^ - 1 - 1 1 1 1 1 = 1 16 2 i ^ + 2 j ^ × - 2 i ^ + 2 j ^ = k ^ 2 = 1 2