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

यदि रेखा \(\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}\) तथा समतल \(x -2 y - kz =3\) के मध्य कोण \(\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\) हो, तो \(k\) का मान होगा

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

Correct Answer

(A) \(\sqrt {\frac{5}{3}} \)

Step-by-step Solution

Detailed explanation

Direction Ratio of line are \(2,1,-2\) Normal vector of plane is \(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\mathrm{k} \hat{\mathrm{k}}\) \(\sin \alpha=\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(\hat{i}-2 \hat{j}-k \hat{k})}{3 \sqrt{1+4+k^{2}}}\)…
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