Let `vecV=2hati+hatj-hatk` and `vecW=hati+3hatk`. It `vecU` is a unit vector, then the maximum value of the scalar triple product `[(vecU, vecV, vecW)]` is

Let vecV = 2hati +hatj - hatk and vecW= hati + 3hatk . if vecU is a unit vector, then the maximum value of the scalar triple product [ vecU vecV vecW] is

Let vecV = 2hati +hatj - hatk and vecW= hati + 3hatk . if vecU is a unit vector, then the maximum value of the scalar triple product [ vecU vecV vecW] is

Let vecV=2hati+hatj-hatk and vecW=hati+3hatk. If vecU is a unit vector then the maximum value of the scalar triple product [vecU vecV vecW] is (A) -1 (B) sqrt(10)+sqrt(6) (C) sqrt(59) (D) sqrt(60)

Let vecv = 2hati + 2hatj - hatk and vecw = hati + 3hatk. If vecu is a unit vector, then the maximum value of the scalar triple product [vecu vecv vecw] is :

Let vecu=2hati-hatj+hatk, vecv=-3hatj+2hatk be vectors and vecw be a unit vector in the xy-plane. Then the maximum possible value of |(vecu xx vecv)|.|vecw| is

Let vecu=2hati-hatj+hatk, vecv=-3hatj+2hatk be vectors and vecw be a unit vector in the xy-plane. Then the maximum possible value of |(vecu xx vecv)|.|vecw| is

Let vecu=2hati-hatj+hatk, vecv=-3hatj+2hatk be vectors and vecw be a unit vector in the xy-plane. Then the maximum possible value of |(vecu xx vecv)|.|vecw| is

Let vecu=hati+hatj,vecv=hati-hatj and vecw=hati+2hatj+3hatk . If hatn is a unit vector such that vecu.hatn=0 and vecv.hatn=0, |vecw.hatn| is equal to (A) 0 (B) 1 (C) 2 (D) 3

Let vecu=hati+hatj,vecv=hati-hatj and vecw=hati+2hatj+3hatk . If hatn is a unit vector such that vecu.hatn=0 and vecv.hatn=0, |vecw.hatn| is equal to (A) 0 (B) 1 (C) 2 (D) 3