The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are d1,d2,d3. Then where K is

Engineering

Mathematics

Direction Cosines and Direction Ratio

theorem in space

Distance from a Plane

Question

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are d₁,d₂,d₃. Then $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = {kd}^{2}$ where K is

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Let the component of the line PQ on x-axis, y-axis and z-axis be dx,dy and dz respectively.

Now, let the angles made by the line with x-axis, y-axis and z-axis be α_1 β4γ respectively.

$∴ d = \sqrt{(dx)^{2} + (dy)^{2} + (dz)^{2}}$

Now, projection of the line on xy plane will be

$d_{1} = \sqrt{{dx}^{2} + {dy}^{2}}$

Similarly, the projection of the line on yz plane will be. $d_{2} = \sqrt{{dz}^{2} + {dy}^{2}}$

and the projection of the line on xz plane will be

$d_{3} = \sqrt{d_{x}^{2} + d_{2}^{2}}$

Hence, ${(d_{1})}^{2} + {(d_{2})}^{2} + {(d_{3})}^{2} = d_{x}^{2} + {dy}^{2} + d_{z}^{2} + d^{2} y + {dx}^{2} + d^{2} z$

By comparing,

$= 2 d^{2} x + 2 d^{2} y + 2 d^{2} z$

= 2((dx)2 + (dy)2 + (dz2)

= 2(d²)

with above, we get k=2

$∴ d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 2 d^{2}$

By comparing $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = {kd}^{2}$ with above, we get

k = 2