A=(4j-7k) cm and B=(5i 3j)cm find sin angel b/w vector
Answers
Answer:
Explanation:
v = 2i + 4j and w = i + 5j then v . w = (2)(1) + (4 )(5) = 22. Exercise Find the dot product of ... The Angle Between Two Vectors ... v = 2i + 3j + k and ... -40 cos 75 i - 40 sin 75 j = -10.3 i - 38.6 j
The unit vectors perpendicular to vectors A and B is
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577k
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577k
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetails
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and B
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8k
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577k
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577kThe other vector perpendicular to A and B is
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577kThe other vector perpendicular to A and B isB x A = -8i +8j +8k
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577kThe other vector perpendicular to A and B isB x A = -8i +8j +8kAgain the magnitude of this vector is
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577kThe other vector perpendicular to A and B isB x A = -8i +8j +8kAgain the magnitude of this vector is√192
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577kThe other vector perpendicular to A and B isB x A = -8i +8j +8kAgain the magnitude of this vector is√192And this unit vector is
The unit vectors perpendicular to vectors A and B is0.577i -0.577j - 0.577kand its negative-0.577i +0.577j + 0.577kDetailsTake the cross product of A and B gives a vector perpendicular to A and BA x B = 8i -8j -8kThe magnitude of this vector is√[8²+8² +8²] = √192 = 13.856The unit vector perpendicular to A and B is thus(1/√192)(8i - 8j - 8k) = 0.577i - 0.577j -0.577kThe other vector perpendicular to A and B isB x A = -8i +8j +8kAgain the magnitude of this vector is√192And this unit vector is(1/√192)(-8i + 8j + 8k) = - 0.577i + 0.577j +0.577k