The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

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 joining the origin (0, 0) and point (-2, 9) is \(m_1=\frac{9-0}{-2-0}=\frac{-9}{2}\).
Accordingly, the slope of the line perpendicular to the line joining the origin and point (-2, 9) is
\(m_2=\frac{-1}{m_1}=-\frac{1}{\left(-\frac{9}{2}\right)}=\frac{2}{9}\).
Now, the equation of the line passing through point (-2, 9) and having a slope \(m_2\) is
\((y-9)=\frac{2}{9}(x+2)\)
\(9y-81=2x+4\)
\(i.e,2x-9y+85=0\)

The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Solution and Explanation

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