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QJEEM22Mab732: JEE Main 2022 Mathematics - Class 12 JEEM22Mab732 Solution: Let and be the vertices of the rhombus PRQS. If the direction…

Let P(−2,−1,1)\mathrm{P}(-2,-1,1) and Q(5617,4317,11117)\mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α,−1,β\alpha,-1, \beta , where both α\alpha and β\beta are integers of minimum absolute values, then α2+β2\alpha^{2}+\beta^{2} is equal to ____________.

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JEE Main 2022 (Online) 28th July Morning Shift Mathematics - 3D Geometry Question 137 English Explanation

d.r's of RS = < \alpha , - 1,\beta >

d.r's of PQ = < {{90} \over {17}},{{60} \over {17}},{{94} \over {17}} > \, = \, < 45,30,47 >

as PQ and RS are diagonals of rhombus

α(45)+30(−1)+47(β)=0\alpha (45) + 30( - 1) + 47(\beta ) = 0

⇒45α+47β=30\Rightarrow 45\alpha + 47\beta = 30

i.e., α=30−47β45\alpha = {{30 - 47\beta } \over {45}}

for minimum integral value α=−15\alpha = - 15 and β=15\beta = 15

⇒α2+β2=450\Rightarrow {\alpha ^2} + {\beta ^2} = 450 .

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