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(B) Let the vector $\overrightarrow{AB} = l\hat{i} + m\hat{j} + n\hat{k}$.The projection of $\overrightarrow{AB}$ on the $XY$ plane is $\sqrt{l^2 + m^

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$1$

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(B) Let the vector $\overrightarrow{AB} = l\hat{i} + m\hat{j} + n\hat{k}$.The projection of $\overrightarrow{AB}$ on the $XY$ plane is $\sqrt{l^2 + m^2} = \sqrt{15}$,so $l^2 + m^2 = 15$ (Equation $1$).The projection of $\overrightarrow{AB}$ on the $YZ$ plane is $\sqrt{m^2 + n^2} = \sqrt{46}$,so $m^2 + n^2 = 46$ (Equation $2$).The projection of $\overrightarrow{AB}$ on the $ZX$ plane is $\sqrt{n^2 + l^2} = 7$,so $n^2 + l^2 = 49$ (Equation $3$).Adding equations $(1)$,$(2)$,and $(3)$:$2(l^2 + m^2 + n^2) = 15 + 46 + 49 = 110$$l^2 + m^2 + n^2 = 55$.To find the projection on the $y$-axis,we need the value of $|m|$.Using $l^2 + m^2 + n^2 = 55$ and $n^2 + l^2 = 49$:$m^2 = 55 - 49 = 6$.Wait,re-evaluating the system:$l^2 + m^2 = 15$$m^2 + n^2 = 46$$n^2 + l^2 = 49$$(l^2 + m^2 + n^2) = (15 + 46 + 49) / 2 = 110 / 2 = 55$.$n^2 = 55 - 15 = 40$.$l^2 = 55 - 46 = 9 \Rightarrow l = 3$.$m^2 = 55 - 49 = 6 \Rightarrow m = \sqrt{6}$.Re-checking the calculation: $l^2+m^2 = 9+6 = 15$ (Correct). $m^2+n^2 = 6+40 = 46$ (Correct). $n^2+l^2 = 40+9 = 49$ (Correct).Therefore,the projection on the $y$-axis is $|m| = \sqrt{6}$.Given the options provided,there might be a typo in the question values. If $l^2+m^2=15, m^2+n^2=46, n^2+l^2=49$ is correct,the answer is $\sqrt{6}$. If the intended answer is $1$,then $m^2=1$.

If the projections of the line segment $\overrightarrow{AB}$ on $XY$,$YZ$,and $ZX$ planes are $\sqrt{15}$,$\sqrt{46}$,and $7$ respectively,then the projection of $\overrightarrow{AB}$ on the $y$-axis is

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