The distance between two points P and Q is d and the length of their projections of P Q on the co-ordinate planes are d1, d2, d3. Then d12+d22+d32=k d2 where ' k ' is

Question

The distance between two points P and Q is d and the length of their projections of P Q on the co-ordinate planes are d1, d2, d3. Then d12+d22+d32=k d2 where ' k ' is (A) 1 (B) 5 (C) 3 (D) 2

Tardigrade Admin · Accepted Answer

Here, d1=d cos (90°-α) ; d2=d cos (90°-β) d3=d cos (90°-&gamma;) ⇒ d12+d22+d32=d2( sin 2 α+ sin 2 β+ sin 2 &gamma;) ⇒ d12+d22+d32=2 d2 ; ∴ k=2

Q. The distance between two points $P$ and $Q$ is $d$ and the length of their projections of $PQ$ on the co-ordinate planes are $d_{1}, d_{2}, d_{3}$ . Then $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}$ where ' $k$ ' is

1

5

3

2

Here, $d_{1} = d cos (9 0^{\circ} - α)$ ;
$d_{2} = d cos (9 0^{\circ} - β)$
$d_{3} = d cos (9 0^{\circ} - γ)$
$\Rightarrow d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = d^{2} (sin^{2} α + sin^{2} β + sin^{2} γ)$
$\Rightarrow d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 2 d^{2}$ ;
$∴ k = 2$

Q. The distance between two points P and Q is d and the length of their projections of PQ on the co-ordinate planes are d1​,d2​,d3​. Then d12​+d22​+d32​=kd2 where ' k ' is

1

5

3

2

Here, d1​=dcos(90∘−α) ; d2​=dcos(90∘−β) d3​=dcos(90∘−γ) ⇒d12​+d22​+d32​=d2(sin2α+sin2β+sin2γ) ⇒d12​+d22​+d32​=2d2; ∴k=2

Q. The distance between two points $P$ and $Q$ is $d$ and the length of their projections of $PQ$ on the co-ordinate planes are $d_{1}, d_{2}, d_{3}$ . Then $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}$ where ' $k$ ' is

Here, $d_{1} = d cos (9 0^{\circ} - α)$ ;
$d_{2} = d cos (9 0^{\circ} - β)$
$d_{3} = d cos (9 0^{\circ} - γ)$
$\Rightarrow d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = d^{2} (sin^{2} α + sin^{2} β + sin^{2} γ)$
$\Rightarrow d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 2 d^{2}$ ;
$∴ k = 2$