Let u, v, w be three vectors such that |u| = | vedclass

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(D) Given $|u| = 1, |v| = 2, |w| = 3$. The projection of $v$ along $u$ is $\frac{v \cdot u}{|u|}$ and the projection of $w$ along $u$ is $\frac{w \cdo

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$\sqrt{14}$

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(D) Given $|u| = 1, |v| = 2, |w| = 3$. The projection of $v$ along $u$ is $\frac{v \cdot u}{|u|}$ and the projection of $w$ along $u$ is $\frac{w \cdot u}{|u|}$.Since projections are equal, $v \cdot u = w \cdot u$, which implies $(v - w) \cdot u = 0$.Also, $v \perp w$, so $v \cdot w = 0$.We need to find $|u - v + w|$. Consider $|u - v + w|^2 = (u - v + w) \cdot (u - v + w)$.$|u - v + w|^2 = |u|^2 + |v|^2 + |w|^2 - 2(u \cdot v) + 2(u \cdot w) - 2(v \cdot w)$.Since $v \cdot u = w \cdot u$, we have $u \cdot v - u \cdot w = 0$.Also $v \cdot w = 0$.$|u - v + w|^2 = 1^2 + 2^2 + 3^2 - 2(u \cdot v - u \cdot w) - 2(0) = 1 + 4 + 9 - 0 - 0 = 14$.Therefore, $|u - v + w| = \sqrt{14}$.

Let $u, v, w$ be three vectors such that $|u| = 1, |v| = 2, |w| = 3$. If the projection of $v$ along $u$ is equal to the projection of $w$ along $u$ and $v, w$ are perpendicular to each other, then $|u - v + w| = ...$

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