Let u, v, w be vectors such that |u| = 1, |v| | vedclass

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(C) Given $|u| = 1, |v| = 2, |w| = 3$. Since $v \perp w$,we have $v \cdot w = 0$.Let the projection of $v$ along $u$ be equal to the projection of $w$

Vedclass · Accepted Answer

$\sqrt{14}$

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(C) Given $|u| = 1, |v| = 2, |w| = 3$. Since $v \perp w$,we have $v \cdot w = 0$.Let the projection of $v$ along $u$ be equal to the projection of $w$ along $u$. This implies $\frac{v \cdot u}{|u|} = \frac{w \cdot u}{|u|}$.Since $|u| = 1$,we have $v \cdot u = w \cdot u$,which means $(v - w) \cdot u = 0$.We want to find $|u - v + w|$. Let $X = u - (v - w)$.Then $|X|^2 = |u - (v - w)|^2 = |u|^2 + |v - w|^2 - 2u \cdot (v - w)$.Since $u \cdot (v - w) = 0$,we have $|X|^2 = |u|^2 + |v - w|^2$.$|v - w|^2 = |v|^2 + |w|^2 - 2(v \cdot w) = 2^2 + 3^2 - 2(0) = 4 + 9 = 13$.Thus,$|u - v + w|^2 = 1^2 + 13 = 14$.Therefore,$|u - v + w| = \sqrt{14}$.

Let $u, v, w$ be 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 $v$ and $w$ are perpendicular to each other,then $|u - v + w|$ equals:

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