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Q 48.

Expert-verifiedFound in: Page 801

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find a vector in the direction opposite to $>">-4,5,-1$ and with magnitude 3.

The required vector is $\frac{3}{\sqrt{42}}>">-4,5,-1$.

The given vector is:

$v=>">-4,5,-1$ with magnitude 3.

$\mathit{v}=>">-4,5,-1$

Therefore, the norm of the given vector is $\sqrt{42}$.

We know that the vector in the direction of **v **is $\frac{a}{||\mathit{v}||}\left(\mathbf{-}\mathit{v}\right)$. Here * a *is the magnitude of the given vector and the negative sign shows the direction opposite to$v=>">-4,5,-1$.

The required vector is:

$\frac{a}{||\mathit{v}||}\mathit{v}=\frac{3}{\sqrt{42}}\left(->">-4,5,-1\right)$

Therefore the vector in the direction opposite to$v=>">-4,5,-1$ and with a magnitude of is $\frac{3}{\sqrt{42}}>">4,-5,1$.

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