JEE Main 2022 — Mathematics Vectors & 3D Geometry

Given,

$P(-2,-1,1)$P(−2,−1,1) and $Q(\frac{56}{17},\frac{43}{17},\frac{111}{17})$Q(1756​,1743​,17111​) be the vertices of the rhombus $PRQS$PRQS. If the direction ratios of the diagonal $RS$RS are $\alpha ,-1,\beta$α,−1,β, where both $\alpha$α and $\beta$β are integers of minimum absolute values, then ${\alpha }^{2}+{\beta }^{2}$α2+β2 is equal to

Now $RS=(\alpha ,-1,\beta )$RS=(α,−1,β)

So, direction ratio of $PQ\equiv (\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1)$PQ≡(1756​+2,1743​+1,17111​−1)

$$\equiv (\frac{90}{17},\frac{60}{17},\frac{94}{17})$$≡(1790​,1760​,1794​)

Now we know that diagonal of rhombus are perpendicular,

So, $\frac{90}{17}\alpha +\frac{60}{17}(-1)+\frac{94}{17}\beta =0$1790​α+1760​(−1)+1794​β=0

$$\Rightarrow 90\alpha +94\beta =60$$⇒90α+94β=60

$$\Rightarrow \beta =\frac{60-90\alpha }{94}$$⇒β=9460−90α​

$$\Rightarrow \beta =\frac{30(2-3\alpha )}{94}$$⇒β=9430(2−3α)​

$$\Rightarrow \beta =-30\frac{(3\alpha -2)}{94}$$⇒β=−3094(3α−2)​

$$\Rightarrow \beta =\frac{-15}{47}(3\alpha -2)$$⇒β=47−15​(3α−2)

$$\Rightarrow \frac{\beta }{-15}=\frac{3\alpha -2}{47}$$⇒−15β​=473α−2​

$\Rightarrow \beta =-15,\alpha =-15$⇒β=−15,α=−15 {\text{as }\alpha \text{ & }\beta \text{ are integer}}

So, ${\alpha }^{2}+{\beta }^{2}=225+225$α2+β2=225+225$=450$=450