Question

If the distance between the points P and Q is d and the projections of PQ on the coordinates planes are ${d_1},{d_2},{d_3},$ Then $d_1^2 + d_2^2 + d_3^2 = $
(A) ${d^2}$
(B) $2{d^2}$
(C) $3{d^2}$
(D) $4{d^2}$

Answer 1

Hint:Projection of a line on a plane is the line joining the point of intersection of the given line with the plane and the foot of the perpendicular from the other point of the given line. Thus a triangle is formed containing the vertical and horizontal components of the given line and the projection of the line.

So, by applying the Pythagoras theorem, we can find out the projection of the given line on
each plane and thus find out the required quantity.

Complete step by step answer:
We are given that the projections of the line PQ of length d on the coordinate planes are
${d_1},{d_2},{d_3}$ .

Let the component of PQ on x-axis be ${d_x}$ , on y-axis be ${d_y}$ and on z-axis be ${d_z}$ . Then,
$d = \sqrt {d_x^2 + d_y^2 + d_z^2} $
Now, we are given that projection of PQ on the XY-plane is ${d_1}$ , so ${d_1} = \sqrt {d_x^2 +
d_y^2} $

Projection of PQ on YZ-plane is ${d_2}$ , so ${d_2} = \sqrt {d_y^2 + d_z^2} $

The projection of PQ on XZ-plane is ${d_3}$ , so ${d_3} = \sqrt {d_x^2 + d_z^2} $
Thus,
$
d_1^2 + d_2^2 + d_3^2 = {(\sqrt {d_x^2 + d_y^2} )^2} + {(\sqrt {d_y^2 + d_z^2} )^2} + {(\sqrt {d_x^2
+ d_z^2} )^2} \\
\Rightarrow d_1^2 + d_2^2 + d_3^2 = d_x^2 + d_y^2 + d_y^2 + d_z^2 + d_x^2 + d_z^2 \\
\Rightarrow d_1^2 + d_2^2 + d_3^2 = 2(d_x^2 + d_y^2 + d_z^2) \\
\Rightarrow d_1^2 + d_2^2 + d_3^2 = 2{d^2} \\
$

Hence, option (B) is the correct answer.

Note:The Pythagoras theorem tells us the relation between the three sides of a right-angled triangle that is the base, the height and the hypotenuse; it states that the sum of the square of the base and height is equal to the square of the hypotenuse. Using this theorem, we find out the values of ${d_1},{d_2},{d_3}$ and thus get the answer to the given question.