First slide

Direction ratios and direction cosines

Question

If $\vec{O P}$ is equally inclined to $\vec{O X}, \vec{O Y} and \vec{O Z}$ and if $P$ is $\sqrt{3}$ units from the origin then $P$ which lies in the first octant is

Easy

A

$(5, 6, 7)$
B

$(2, 3, 4)$
C

$(1, 1, 1)$
D

$(- 1, - 1, 1)$

Solution

If the length of the line $O P = r$ and direction cosines of $\bar{O P}$ are $〈l, m, n〉$ then the coordinates of the point $P (l r, m r, n r)$
Since the line $\vec{O P}$ makes equal angles with coordinate axes, the direction cosines of $\vec{O P}$ are $〈\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}〉$
Therefore, the coordinates of the point $P$
where $O P = \sqrt{3}$ are $P (1, 1, 1)$ take all positive coordinates because the point $P$ lies in the first octant .

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