Find equation of the line through the point (0, 2) making an angle \(\frac{2\pi}{3}\) with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Question

If the components of \( \vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) along and perpendicular to \( \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \) respectively are \( \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) \) and \( \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}) \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

Answer 1

The slope of the line making an angle \(\frac{2\pi}{3}\) with the positive x-axis is \(m=tan\)(\(\frac{2\pi}{3}\))=\(-\sqrt{3}\)

Now, the equation of the line passing through point (0, 2) and having a slope \(-\sqrt{3}\) is\((y-2)=-\sqrt{3}(x-0)\) .
\(y-2=\sqrt{3}x\)
i.e, \(\sqrt3x+y-2=0\)

The slope of line parallel to line \(\sqrt3x+y-2=0\) is \(-\sqrt{3}\)
It is given that the line parallel to line \(\sqrt3x+y-2=0\) crosses the y-axis 2 units below the origin i.e., it passes
through point (0,2).
Hence, the equation of the line passing through the point (0,2) and having a slope \(-\sqrt{3}\) is
\(y-\left(-2\right)=-\sqrt{3}\left(x-0\right)\)
\(y+2=-\sqrt3x\)
\(\sqrt3x+y+2=0\)

Find equation of the line through the point (0, 2) making an angle \(\frac{2\pi}{3}\) with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution and Explanation

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