Question 111Asked in BPSC TRE 1.0 (2023) Mathematics Class 11-12

The minimum distance between the lines (x − 3)/3 = (y − 8)/(−1) = z − 3 and (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4 isरेखाओं (x − 3)/3 = (y − 8)/(−1) = z − 3 और (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4 के बीच की न्यूनतम दूरी है

रेखाओं (x − 3)/3 = (y − 8)/(−1) = z − 3 और (x + 3)/(−3) = (y + 7)/2 = (z − 6)/4 के बीच की न्यूनतम दूरी है
  1. A2√302√30
  2. B√30√30
  3. C3√303√30
  4. DMore than one of the aboveउपर्युक्त में से एक से अधिक
  5. ENone of the aboveउपर्युक्त में से कोई नहीं
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Correct answer: (C) 3√303√30

Explanation

d₁ = (3, −1, 1) and d₂ = (−3, 2, 4), so d₁ × d₂ = (−6, −15, 3) with magnitude √270. The joining vector is (−3, −7, 6) − (3, 8, 3) = (−6, −15, 3). Distance = 270/√270 = √270 = 3√30.d₁ = (3, −1, 1) और d₂ = (−3, 2, 4), इसलिए d₁ × d₂ = (−6, −15, 3), जिसका परिमाण √270 है। मिलाने वाला सदिश (−3, −7, 6) − (3, 8, 3) = (−6, −15, 3) है। दूरी = 270/√270 = √270 = 3√30।

From BPSC TRE 1.0 (2023) Mathematics Class 11-12 — Previous Year Paper

This is question 111 of BPSC TRE 1.0 (2023) Mathematics Class 11-12 — Previous Year Paper. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.