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# Find a direction vector $\mathbf{d} = \begin{pmatrix}d_1 \\ d_2 \\d_3 \end{pmatrix}$for the line through $B = (1, 1, 2)$ and \$C= (2, 3,

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Find a direction vector $$\mathbf{d} = \begin{pmatrix}d_1 \\ d_2 \\d_3 \end{pmatrix}$$

for the line through B = (1, 1, 2) and C= (2, 3, 1) such that $$d_1 + d_2 + d_3 = 10.$$

Mar 14, 2019

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Find a direction vector

$$\mathbf{d} = \begin{pmatrix}d_1 \\ d_2 \\d_3 \end{pmatrix}$$
for the line through

$$B = (1, 1, 2)$$ and $$C= (2, 3, 1)$$

such that

$$d_1 + d_2 + d_3 = 10$$.

Line:

$$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{B} + \underbrace{\lambda \left( \vec{C} - \vec{B} \right) }_{=\vec{d}} \\\\ \vec{d} &=& \lambda \left( \vec{C} - \vec{B} \right) \\ \vec{d} &=& \lambda \begin{pmatrix}2-1 \\ 3-1 \\1-2 \end{pmatrix} \\ \vec{d} &=& \lambda \begin{pmatrix}1 \\ 2 \\ -1 \end{pmatrix} \\ \begin{pmatrix}d_1 \\ d_2 \\d_3 \end{pmatrix} &=& \lambda \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \\\\ && d_1 = \lambda \\ && d_2 =2\lambda \\ && d_3 = -\lambda \\ \hline && d_1+d_2+d_3 = \lambda+2\lambda -\lambda \\ && 10 = 2\lambda \\ && \lambda =\dfrac{10}{2} \\ && \lambda = 5 \\\\ \vec{d} &=& 5 \begin{pmatrix}1 \\ 2 \\ -1 \end{pmatrix} \\ \mathbf{\vec{d}} & \mathbf{=} & \mathbf{ \begin{pmatrix} 5 \\ 10 \\ -5 \end{pmatrix} } \\ \hline \end{array}$$

Mar 15, 2019