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

ધારોકે સમતલ \(x+3 y-2 z+6=0\) યામાક્ષોને બિંદુુો \(A, B, C\) પાસે મળે છે.જો ત્રિકોણ \(ABC\) નું લંબકેન્દ્ર \(\left(\alpha, \beta, \frac{6}{7}\right)\) હોય, તો \(98(\alpha+\beta)^2=...........\)

  1. A \(280\)
  2. B \(281\)
  3. C \(282\)
  4. D \(288\)
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Answer & Solution

Correct Answer

(D) \(288\)

Step-by-step Solution

Detailed explanation

\(A (-6,0,0) \quad B (0,-2,0) \quad C =(0,0,3)\) \(\overrightarrow{ AB }=6 \hat{ i }-2 \hat{ j }, \quad \overrightarrow{ BC }=2 \hat{ j }+3 \hat{ k }\), \(\overrightarrow{ AC }=6 \hat{ i }+3 \hat{ k }\) \(\overrightarrow{ AH } \cdot \overrightarrow{ BC }=0\)…
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