$A B = d$ and $d_{1}, d_{2}, d_{3}$ are the projections of the llne segment $A B$ on the coordinate planes, then $d^{2} =$

\frac{d_{1}^{2} + d_{2}^{2} + d_{3}^{2}}{2}

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\frac{d_{1} + d_{2} + d_{3}}{3}

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\frac{d_{1} d_{2} d_{3}}{2}

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2 (d_{1}^{2} + d_{2}^{2} + d_{3}^{2})

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Solution

The correct option is A $\frac{d_{1}^{2} + d_{2}^{2} + d_{3}^{2}}{2}$
Let the components of the line $A B$ on $x$ axis, $y$ axis and $z$ axis be $d_{x}$ $d_{y}$ and $d_{z}$ respectively.
Now, let the angles made by the line with $x$ axis, $y$ axis and $z$ axis be $α$ $β$ and $γ$ respectively.
Therefore $d = \sqrt{(d_{x})^{2} + (d_{x})^{2} + (d_{x})^{2}}$
Now the projection of the line on xy plane will be $d_{1} = \sqrt{d_{x}^{2} + d_{y}^{2}}$
Similarly, the projection of the line on $y z$ plane will be $d_{2} = \sqrt{d_{z}^{2} + d_{y}^{2}}$
and the projection of the line on $x z$ plane will be $d_{3} = \sqrt{d_{x}^{2} + d_{z}^{2}}$
Hence, $(d_{1})^{2} + (d_{2})^{2} + (d_{3})^{2}$
$= d_{x}^{2} + d_{y}^{2} + d_{z}^{2} + d_{y}^{2} + d_{x}^{2} + d_{z}^{2}$
$= 2 ((d_{x})^{2} + (d_{x})^{2} + (d_{x})^{2})$
$= 2 (d^{2})$
Hence, $d^{2} = \frac{(d_{1})^{2} + (d_{2})^{2} + (d_{3})^{2}}{2}$

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