a, b and c are three vectors such that |a|=1, | vedclass

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(C) Given that,$|a|=1, |b|=2, |c|=3$. Since $b$ and $c$ are perpendicular,we have $b \cdot c = 0$. The projection of $b$ on $a$ is $\frac{a \cdot b}{|

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$\sqrt{14}$

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(C) Given that,$|a|=1, |b|=2, |c|=3$. Since $b$ and $c$ are perpendicular,we have $b \cdot c = 0$. The projection of $b$ on $a$ is $\frac{a \cdot b}{|a|}$ and the projection of $c$ on $a$ is $\frac{a \cdot c}{|a|}$. Given that these projections are equal,we have $\frac{a \cdot b}{|a|} = \frac{a \cdot c}{|a|}$,which implies $a \cdot b = a \cdot c$. Now,we calculate $|a-b+c|^2$: $|a-b+c|^2 = |a|^2 + |b|^2 + |c|^2 - 2(a \cdot b) + 2(a \cdot c) - 2(b \cdot c)$. Substituting the known values: $|a-b+c|^2 = (1)^2 + (2)^2 + (3)^2 - 2(a \cdot b) + 2(a \cdot b) - 2(0)$. $|a-b+c|^2 = 1 + 4 + 9 = 14$. Therefore,$|a-b+c| = \sqrt{14}$.

$a, b$ and $c$ are three vectors such that $|a|=1, |b|=2, |c|=3$ and $b, c$ are perpendicular. If the projection of $b$ on $a$ is the same as the projection of $c$ on $a$,then $|a-b+c|$ is equal to

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