The distance between two points P and Q is d | vedclass

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Question

(D) Let the coordinates of $P$ be $(x_1, y_1, z_1)$ and $Q$ be $(x_2, y_2, z_2)$.Let $\Delta x = x_2 - x_1$,$\Delta y = y_2 - y_1$,and $\Delta z = z_2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Let the coordinates of $P$ be $(x_1, y_1, z_1)$ and $Q$ be $(x_2, y_2, z_2)$.Let $\Delta x = x_2 - x_1$,$\Delta y = y_2 - y_1$,and $\Delta z = z_2 - z_1$.The distance $d$ is given by $d^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$.The projections of $PQ$ on the $xy$,$yz$,and $zx$ planes are $d_1, d_2, d_3$ respectively.Thus,$d_1^2 = (\Delta x)^2 + (\Delta y)^2$,$d_2^2 = (\Delta y)^2 + (\Delta z)^2$,and $d_3^2 = (\Delta z)^2 + (\Delta x)^2$.Adding these,we get $d_1^2 + d_2^2 + d_3^2 = 2((\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2) = 2d^2$.Comparing this with $d_1^2 + d_2^2 + d_3^2 = k d^2$,we find $k = 2$.

The distance between two points $P$ and $Q$ is $d$ and the lengths of the projections of $PQ$ on the coordinate planes are $d_1, d_2, d_3$. Then $d_1^2 + d_2^2 + d_3^2 = k d^2$,where $k$ is:

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