(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6) %3D с(х, у, 2) %3D (сх + бс 6, су + бс - 6, сz + бс б) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.
(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6) %3D с(х, у, 2) %3D (сх + бс 6, су + бс - 6, сz + бс б) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.
(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6) %3D с(х, у, 2) %3D (сх + бс 6, су + бс - 6, сz + бс б) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.
Rather than use the standard definitions of addition and scalar multiplication in R3, suppose these two operations are defined as follows. With these new definitions, is R3, a vector space? Justify your answers.
Transcribed Image Text:(X1, Y1, Z1) + (x2, Y2, Z2) = (x1 + X2 + 6, y1 + Y2 + 6, Z1 + Z2 + 6)
(p).
c(x, y, z) = (cx + 6c – 6, cy + 6c - 6, cz + 6c – 6)
The set is a vector space.
O The set is not a vector space because the additive identity property is not satisfied.
O The set is not a vector space because it is not closed under scalar multiplication.
O The set is not a vector space because the distributive property is not satisfied.
O The set is not a vector space because the multiplicative identity property is not satisfied.
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
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