To solve the given vector operations, we will:
To find 23u\frac{2}{3} \mathbf{u}32u, we multiply each component of u\mathbf{u}u by 23\frac{2}{3}32: u=⟨7,9⟩ \mathbf{u} = \langle 7, 9 \rangle u=⟨7,9⟩ 23u=⟨23⋅7,23⋅9⟩=⟨4.6667,6⟩ \frac{2}{3} \mathbf{u} = \left\langle \frac{2}{3} \cdot 7, \frac{2}{3} \cdot 9 \right\rangle = \left\langle 4.6667, 6 \right\rangle 32u=⟨32⋅7,32⋅9⟩=⟨4.6667,6⟩
To find 3v3 \mathbf{v}3v, we multiply each component of v\mathbf{v}v by 3: v=⟨6,−6⟩ \mathbf{v} = \langle 6, -6 \rangle v=⟨6,−6⟩ 3v=⟨3⋅6,3⋅−6⟩=⟨18,−18⟩ 3 \mathbf{v} = \left\langle 3 \cdot 6, 3 \cdot -6 \right\rangle = \left\langle 18, -18 \right\rangle 3v=⟨3⋅6,3⋅−6⟩=⟨18,−18⟩
To find v−u\mathbf{v} - \mathbf{u}v−u, we subtract each component of u\mathbf{u}u from the corresponding component of v\mathbf{v}v: v−u=⟨6−7,−6−9⟩=⟨−1,−15⟩ \mathbf{v} - \mathbf{u} = \left\langle 6 - 7, -6 - 9 \right\rangle = \left\langle -1, -15 \right\rangle v−u=⟨6−7,−6−9⟩=⟨−1,−15⟩
To find 2u+5v2 \mathbf{u} + 5 \mathbf{v}2u+5v, we first multiply each component of u\mathbf{u}u by 2 and each component of v\mathbf{v}v by 5, then add the corresponding components: 2u=⟨2⋅7,2⋅9⟩=⟨14,18⟩ 2 \mathbf{u} = \left\langle 2 \cdot 7, 2 \cdot 9 \right\rangle = \left\langle 14, 18 \right\rangle 2u=⟨2⋅7,2⋅9⟩=⟨14,18⟩ 5v=⟨5⋅6,5⋅−6⟩=⟨30,−30⟩ 5 \mathbf{v} = \left\langle 5 \cdot 6, 5 \cdot -6 \right\rangle = \left\langle 30, -30 \right\rangle 5v=⟨5⋅6,5⋅−6⟩=⟨30,−30⟩ 2u+5v=⟨14+30,18+−30⟩=⟨44,−12⟩ 2 \mathbf{u} + 5 \mathbf{v} = \left\langle 14 + 30, 18 + -30 \right\rangle = \left\langle 44, -12 \right\rangle 2u+5v=⟨14+30,18+−30⟩=⟨44,−12⟩
(a) 23u=⟨4.6667,6⟩\frac{2}{3} \mathbf{u} = \boxed{\left\langle 4.6667, 6 \right\rangle}32u=⟨4.6667,6⟩
(b) 3v=⟨18,−18⟩3 \mathbf{v} = \boxed{\left\langle 18, -18 \right\rangle}3v=⟨18,−18⟩
(c) v−u=⟨−1,−15⟩\mathbf{v} - \mathbf{u} = \boxed{\left\langle -1, -15 \right\rangle}v−u=⟨−1,−15⟩
(d) 2u+5v=⟨44,−12⟩2 \mathbf{u} + 5 \mathbf{v} = \boxed{\left\langle 44, -12 \right\rangle}2u+5v=⟨44,−12⟩
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