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Express the vector $\vec{u} = 2\hat{i}+4\hat{j}+5\hat{k}$ as a sum of a vector $\vec{a}$ parallel to $\vec{v}=2\hat{i}-\hat{j}-2\hat{k}$ and a vector $\vec{b}$ perpendicular to $\vec{v}$.

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  • $\begingroup$ It should be good if you write what you have done. $\endgroup$
    – Sigur
    Jul 4 '12 at 21:18
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    $\begingroup$ The conditions shown yield $\vec{b}\cdot\vec{v} = 0$, $\vec{u} = \vec{a}+\vec{b}$, and $\vec{a} = k{\vec{v}}$. This is enough to go about it algebraically. $\endgroup$ Jul 4 '12 at 21:21
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Projection of vector $\vec u$ along vector $\vec v= \vec u.\vec v/|\vec v|= -10/3$ and thus the part of vector $\vec u$ along vector $\vec v=(-10/3)(2 \vec i -\vec j -2\vec k/3)$(unit vector along $\vec v$) and thus the part of vector $\vec u$ perpendicular to $\vec v= \vec u-(-10/9(2 \vec i -\vec j -2\vec k))$.

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Just to give what I see as a slightly more elegant take on the same spiel, note that by $\vec u = \vec a + \vec b$ and $\vec a= k \vec v$, we have a vector equality $\vec u = k \vec v + \vec b$. Taking a dot product on both sides by $\vec v$ yields $$ \vec u \cdot \vec v = k|v|^2 + \vec b \cdot \vec v~~. $$ The orthogonality condition makes the last term vanish, yielding $k = \dfrac{\vec u \cdot \vec v}{|v|^2}$ , from which the solution can be quickly deduced.

To wit, in this case $k = \frac{-10}{9}$ , and so $\vec a = \frac{1}{9}<-20,10,20>$ and $ \vec b = \frac{1}{9} <38,26,25> $.

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