JEE Mains · Maths · STD 12 - 10. vector algebra
Let A, B, C be three points in \(x y\)-plane, whose position vector are given by \(\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}\) and \(\mathrm{a} \hat{i}+(1-\mathrm{a}) \hat{j}\) respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) is \(\frac{9}{\sqrt{2}}\), then the sum of all the possible values of \(a\) is :
- A \(2\)
- B \(9 / 2\)
- C \(1\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
Equation of line in the internal bisector of \(O \dot{A}\) and \(O B\) is \((\sqrt{3}+1) \hat{i}+(\sqrt{3}+1) \hat{j}\) \(\Rightarrow\) line will be \(y=x \Rightarrow x-y=0\) \(D=\left|\frac{a-(1-a)}{\sqrt{a^2+(1-a)^2}}\right|=\frac{9}{\sqrt{2}}\)…
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