Question

Let P(–2, –1, 1) and Q(56/17, 43/17, 111/17) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are \(α, –1, β,\) where both \(α\) and \(β\) are integers of minimum absolute values, then \(α^2 + β^2\) is equal to ___________.

Answer 1

\(RS≡(α,−1,β)\)

\(DR \space of\space  PQ\)≡\((\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}−1)\)
≡\((\frac{90}{17},\frac{60}{17},\frac{94}{17})\)

\(\frac{90}{17}α+\frac{60}{17}(−1)+\frac{94}{17}β=0\)

\(90α+94β=60\)
\(β=\frac{60−90α}{94}\)

\(β=\frac{30(2−3α)}{94}\)

\(β=−30\frac{(3α−2)}{94}\)

\(β=−\frac{15}{47}(3α−2)\)

\(⇒\frac{β}{-15}=\frac{3α−2}{47}\)

\(⇒β=−15,α=−15\)

\(α^2+β^2=225+225\)

\(=450\)