JEE Mains · Maths · STD 12 - 11. three dimension geometry
If an angle between the line, \(\frac{{x + 1}}{2} = \frac{{y - 2}}{1} = \frac{{z - 3}}{{ - 2}}\) and the plane, \(x\,\, - \,2y\, - \,kz\, = \,3\) is \({\cos ^{ - 1}}\,\left( {\frac{{2\sqrt 2 }}{3}} \right),\) then a value of \(k\) is
- A \(\sqrt {\frac{5}{3}} \)
- B \(\sqrt {\frac{3}{5}} \)
- C \( - \frac{3}{5}\)
- D \( - \frac{5}{3}\)
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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