Consider the vectors vec{a}=3 hat{i}+5 hat{j} | vedclass

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(B) The length of the projection of vector $\vec{u}$ on vector $\vec{v}$ is given by $\frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|}$.$l = \frac{|\vec{a} \

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$l=m=n$

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(B) The length of the projection of vector $\vec{u}$ on vector $\vec{v}$ is given by $\frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|}$.$l = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \frac{|(3)(2) + (5)(-3) + (2)(-5)|}{\sqrt{2^2 + (-3)^2 + (-5)^2}} = \frac{|6 - 15 - 10|}{\sqrt{4 + 9 + 25}} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$.$m = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|} = \frac{|(2)(-5) + (-3)(-2) + (-5)(3)|}{\sqrt{(-5)^2 + (-2)^2 + 3^2}} = \frac{|-10 + 6 - 15|}{\sqrt{25 + 4 + 9}} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$.$n = \frac{|\vec{c} \cdot \vec{a}|}{|\vec{a}|} = \frac{|(-5)(3) + (-2)(5) + (3)(2)|}{\sqrt{3^2 + 5^2 + 2^2}} = \frac{|-15 - 10 + 6|}{\sqrt{9 + 25 + 4}} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$.Since $l = \frac{19}{\sqrt{38}}$, $m = \frac{19}{\sqrt{38}}$, and $n = \frac{19}{\sqrt{38}}$, we have $l = m = n$.

Consider the vectors $\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}$, $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{c}=-5 \hat{i}-2 \hat{j}+3 \hat{k}$. If $l, m$ and $n$ are the lengths of the projections of $\vec{a}$ on $\vec{b}$, $\vec{b}$ on $\vec{c}$ and $\vec{c}$ on $\vec{a}$ respectively, then:

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