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Let a=(1,0,1)\mathbf{a}=(1,0,1)a=(1,0,1), b=(3,2,1)\mathbf{b}=(3,2,1)b=(3,2,1), and c=(5,6,1)\mathbf{c}=(5,6,1)c=(5,6,1). Compute (a+b)×c(\mathbf{a}+\mathbf{b})\times \mathbf{c}(a+b)×c.
Let a=(1,0,1)\mathbf{a}=(1,0,1)a=(1,0,1), b=(3,2,1)\mathbf{b}=(3,2,1)b=(3,2,1), and c=(5,6,1)\mathbf{c}=(5,6,1)c=(5,6,1). Compute (b−c)×a(\mathbf{b}-\mathbf{c})\times \mathbf{a}(b−c)×a.
Let a=(101)\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}a=101, b=(321)\mathbf{b} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}b=321, and c=(561)\mathbf{c} = \begin{pmatrix} 5 \\ 6 \\ 1 \end{pmatrix}c=561.
Calculate (2a+3b)×(b−c)(2\mathbf{a} + 3\mathbf{b}) \times (\mathbf{b} - \mathbf{c})(2a+3b)×(b−c).
Let a=(1,0,1)\mathbf{a}=(1,0,1)a=(1,0,1) and c=(5,6,1)\mathbf{c}=(5,6,1)c=(5,6,1). Calculate c×a\mathbf{c}\times \mathbf{a}c×a.
Let a=(101)\mathbf{a}=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}a=101, b=(321)\mathbf{b}=\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}b=321, and c=(561)\mathbf{c}=\begin{pmatrix} 5 \\ 6 \\ 1 \end{pmatrix}c=561. Calculate a×b\mathbf{a}\times \mathbf{b}a×b.
Let a=(101)\mathbf{a}=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}a=101, b=(321)\mathbf{b}=\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}b=321, and c=(561)\mathbf{c}=\begin{pmatrix} 5 \\ 6 \\ 1 \end{pmatrix}c=561. Verify the distributive law by simplifying (a+b)×(b+c)(\mathbf{a}+\mathbf{b})\times(\mathbf{b}+\mathbf{c})(a+b)×(b+c) and comparing with a×b+a×c+b×b+b×c\mathbf{a}\times \mathbf{b} + \mathbf{a}\times \mathbf{c} + \mathbf{b}\times \mathbf{b} + \mathbf{b}\times \mathbf{c}a×b+a×c+b×b+b×c.
Let a=(1,0,1)a=(1,0,1)a=(1,0,1), b=(3,2,1)b=(3,2,1)b=(3,2,1), and c=(5,6,1)c=(5,6,1)c=(5,6,1). Compute b×cb\times cb×c.
Let a=(1,0,1)\mathbf{a}=(1,0,1)a=(1,0,1), b=(3,2,1)\mathbf{b}=(3,2,1)b=(3,2,1), and c=(5,6,1)\mathbf{c}=(5,6,1)c=(5,6,1). Compute the scalar triple product a⋅(b×c)\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})a⋅(b×c).
Let a=(101)\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}a=101, b=(321)\mathbf{b} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}b=321, and c=(561)\mathbf{c} = \begin{pmatrix} 5 \\ 6 \\ 1 \end{pmatrix}c=561. Calculate a×(b+c)\mathbf{a} \times (\mathbf{b} + \mathbf{c})a×(b+c).
Let a=(1,0,1)\mathbf{a}=(1,0,1)a=(1,0,1), b=(3,2,1)\mathbf{b}=(3,2,1)b=(3,2,1), and c=(5,6,1)\mathbf{c}=(5,6,1)c=(5,6,1). Compute (a×b)+(b×c)+(c×a)(\mathbf{a}\times \mathbf{b})+(\mathbf{b}\times \mathbf{c})+(\mathbf{c}\times \mathbf{a})(a×b)+(b×c)+(c×a)
Let a=(101)\mathbf{a}=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}a=101, b=(321)\mathbf{b}=\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}b=321, and c=(561)\mathbf{c}=\begin{pmatrix} 5 \\ 6 \\ 1 \end{pmatrix}c=561. Find a unit vector perpendicular to both (a+b)(\mathbf{a}+\mathbf{b})(a+b) and (b+c)(\mathbf{b}+\mathbf{c})(b+c).
Let a=(101)\mathbf{a}=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}a=101, b=(321)\mathbf{b}=\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}b=321, and c=(561)\mathbf{c}=\begin{pmatrix} 5 \\ 6 \\ 1 \end{pmatrix}c=561. Define u=a+2c\mathbf{u}=\mathbf{a}+2\mathbf{c}u=a+2c and v=3b−a\mathbf{v}=3\mathbf{b} - \mathbf{a}v=3b−a.
Calculate u×v\mathbf{u}\times \mathbf{v}u×v.
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