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JEE Mains · Maths · STD 12 - 11. three dimension geometry

જો દિગ્ગુણોત્તર \(3, – 1, 0\) વાળી એક રેખાની દિશામાં, બિંદુ \(P (43, \alpha, \beta), \beta<0\) નુ રેખા \(\vec{ r }=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in R\) થી અંતર \(13 \sqrt{10}\) હોય, તો \(\alpha^2+\beta^2=\) ___ .

  1. A 170
  2. B 160
  3. C 180
  4. D 150
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Correct Answer

(A) 170

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\(\frac{x-43}{3}=\frac{y-\alpha}{-1}=\frac{z-\beta}{0} \Rightarrow P_1(43+3 \lambda, \alpha-\lambda, \beta)\) \(\frac{ x -4}{2}=\frac{ y }{0}=\frac{ z +1}{3} \Rightarrow P _1(2 \mu+4,0,3 \mu-1)\)…
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