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(A) Let the coordinates of $A$ be $(x_1, y_1, z_1)$ and $B$ be $(x_2, y_2, z_2)$. The distance $d$ is given by $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 +

Vedclass · Accepted Answer

$2 d^2 = d_1^2 + d_2^2 + d_3^2$

Vedclass · Answer

(A) Let the coordinates of $A$ be $(x_1, y_1, z_1)$ and $B$ be $(x_2, y_2, z_2)$. The distance $d$ is given by $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$. Let $\Delta x = x_2 - x_1$,$\Delta y = y_2 - y_1$,and $\Delta z = z_2 - z_1$. Then $d^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$. The projection of $AB$ on the $XY$-plane has length $d_1 = \sqrt{(\Delta x)^2 + (\Delta y)^2}$. The projection of $AB$ on the $YZ$-plane has length $d_2 = \sqrt{(\Delta y)^2 + (\Delta z)^2}$. The projection of $AB$ on the $ZX$-plane has length $d_3 = \sqrt{(\Delta z)^2 + (\Delta x)^2}$. Squaring these,we get $d_1^2 = (\Delta x)^2 + (\Delta y)^2$,$d_2^2 = (\Delta y)^2 + (\Delta z)^2$,and $d_3^2 = (\Delta z)^2 + (\Delta x)^2$. Adding these three equations: $d_1^2 + d_2^2 + d_3^2 = 2((\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2)$. Since $d^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$,we have: $d_1^2 + d_2^2 + d_3^2 = 2d^2$.

If the distance between two points $A$ and $B$ is $d$,and the lengths of the projections of $AB$ on the coordinate planes are $d_1, d_2, d_3$,then

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