Let u = i + j,v = i - j and w = i + 2j + 3k. | vedclass

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(D) Given vectors are $u = i + j$ and $v = i - j$. Since $n$ is a unit vector perpendicular to both $u$ and $v$,it must be parallel to the cross produ

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$3$

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(D) Given vectors are $u = i + j$ and $v = i - j$. Since $n$ is a unit vector perpendicular to both $u$ and $v$,it must be parallel to the cross product $u 	imes v$. First,calculate the cross product: $u 	imes v = \begin{vmatrix} i & j & k \ 1 & 1 & 0 \ 1 & -1 & 0 \end{vmatrix} = i(0) - j(0) + k(-1 - 1) = -2k$. The magnitude of this vector is $|-2k| = 2$. Thus,the unit vector $n$ is $\pm \frac{-2k}{2} = \pm (-k)$. Now,calculate $|w \cdot n|$ where $w = i + 2j + 3k$: $|w \cdot n| = |(i + 2j + 3k) \cdot (\pm k)| = |\pm 3| = 3$.

Let $u = i + j$,$v = i - j$ and $w = i + 2j + 3k$. If $n$ is a unit vector such that $u \cdot n = 0$ and $v \cdot n = 0$,then $|w \cdot n|$ is equal to

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