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If the lengths of projections of a line of length $l$ over the coordinate axes are $l_1, l_2$ and $l_3$ respectively, then $l_1^2+l_2^2+l_3^2$ is equal to
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The correct answer is:
$l^2$
It d be the length of line, then projection on X-axis = dl, projection on Y-axis = dm
and projection on Z-axis = dn
$\begin{aligned} & \text { Given, } d=l \text { and } d l=l_1 \\ & \qquad \begin{array}{l}d m=l_2 \\ \text { and } d n=l_3\end{array}\end{aligned}$
$\begin{aligned} & \text { Now, }(d l)^2+(d m)^2+(d n)^2=l_1^2+l_2^2+l_3^2 \\ & d^2\left(l^2+m^2+n^2\right)=l_1^2+l_2^2+l_3^2\left\{\because l^2+m^2+n^2=1\right\} \\ & l^2 \cdot l=l_1^2+l_2^2+l_3^2 \Rightarrow l_1^2+l_2^2+l_3^2=l^2\end{aligned}$
and projection on Z-axis = dn
$\begin{aligned} & \text { Given, } d=l \text { and } d l=l_1 \\ & \qquad \begin{array}{l}d m=l_2 \\ \text { and } d n=l_3\end{array}\end{aligned}$
$\begin{aligned} & \text { Now, }(d l)^2+(d m)^2+(d n)^2=l_1^2+l_2^2+l_3^2 \\ & d^2\left(l^2+m^2+n^2\right)=l_1^2+l_2^2+l_3^2\left\{\because l^2+m^2+n^2=1\right\} \\ & l^2 \cdot l=l_1^2+l_2^2+l_3^2 \Rightarrow l_1^2+l_2^2+l_3^2=l^2\end{aligned}$
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