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

यदि समतलों \(a x+b y=3\) तथा \(a x+b y+c z=0\), \(a > 0\) की प्रतिच्छेदन रेखा, समतल \(y - z +2=0\) से \(30^{\circ}\) का कोण बनाती है, तो रेखा की दिक् कोज्या है:

  1. A \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\)
  2. B \(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\)
  3. C \(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}, 0\)
  4. D \(A\) or \(B\) or both
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Answer & Solution

Correct Answer

(D) \(A\) or \(B\) or both

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\(\begin{array}{l}\overrightarrow{ n }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } &\hat{ k } \\a & b & 0 \\a & b & c\end{array}\right| \\= bc \hat{ i }- ac \hat {j }\end{array}\) Direction ratios of line are \(b ,- a , 0\) Direction ratios of normal of the plane…
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