# How do you write the vector a = 3i + 2j - 6k as a sum of two vectors, one parallel and one perpendicular to d = 2i - 4j + k?

$\lambda d = - \frac{8}{21} \left(2 , - 4 , 1\right)$

$n = \frac{1}{21} \left(79 , 10 , - 118\right)$

#### Explanation:

a = lambda d + n ; and n * d = 0
a = ((3), (2), (-6)) ; d = ((2), (-4), (1)) ; n = ((x), (y), (z))

$n \cdot d = 2 x - 4 y + z = 0 R i g h t a r r o w \frac{2 x + z}{4} = y$

$n = \left(\begin{matrix}x \\ \frac{x}{2} + \frac{z}{4} \\ z\end{matrix}\right) = x \cdot \left(\begin{matrix}1 \\ \frac{1}{2} \\ 0\end{matrix}\right) + z \cdot \left(\begin{matrix}0 \\ \frac{1}{4} \\ 1\end{matrix}\right)$

$a = \lambda d + n \Leftrightarrow \left(\begin{matrix}3 \\ 2 \\ - 6\end{matrix}\right) = \lambda \cdot \left(\begin{matrix}2 \\ - 4 \\ 1\end{matrix}\right) + x \cdot \left(\begin{matrix}1 \\ \frac{1}{2} \\ 0\end{matrix}\right) + z \cdot \left(\begin{matrix}0 \\ \frac{1}{4} \\ 1\end{matrix}\right)$

$\left(\begin{matrix}3 \\ 8 \\ - 6\end{matrix}\right) = \left(\begin{matrix}2 & 1 & 0 \\ - 16 & 2 & 1 \\ 1 & 0 & 1\end{matrix}\right) \cdot \left(\begin{matrix}\lambda \\ x \\ z\end{matrix}\right)$

${L}_{2} : = {L}_{2} - {L}_{3}$

$\left(\begin{matrix}3 \\ 14 \\ - 6\end{matrix}\right) = \left(\begin{matrix}2 & 1 & 0 \\ - 17 & 2 & 0 \\ 1 & 0 & 1\end{matrix}\right) \cdot \left(\begin{matrix}\lambda \\ x \\ z\end{matrix}\right)$

${L}_{1} : = {L}_{2} - 2 {L}_{1}$

$\left(\begin{matrix}8 \\ 14 \\ - 6\end{matrix}\right) = \left(\begin{matrix}- 21 & 0 & 0 \\ - 17 & 2 & 0 \\ 1 & 0 & 1\end{matrix}\right) \cdot \left(\begin{matrix}\lambda \\ x \\ z\end{matrix}\right)$

$\lambda = - \frac{8}{21}$

$14 = 17 \cdot \frac{8}{21} + 2 x R i g h t a r r o w \frac{14 \cdot 21 - 17 \cdot 8}{42} = x = \frac{79}{21}$

$- 6 = - \frac{8}{21} + z R i g h t a r r o w \frac{- 6 \cdot 21 + 8}{21} = z = - \frac{118}{21}$

$y = \frac{1}{4} \left(2 x + z\right) = \frac{1}{4} \cdot \frac{2 \cdot 79 - 118}{21} = \frac{10}{21}$