Question

The coordinates of the point $P \equiv (1,2,3)$ and $O \equiv (0,0,0)$ are given. The direction cosines of $\overline{OP}$ are

• A. $\frac{1}{\sqrt{14}},\ \frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}}$
• B. $\frac{1}{\sqrt{6}},\ \frac{2}{\sqrt{6}},\ \frac{1}{\sqrt{6}}$
• C. $\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}$
• D. $\frac{2}{\sqrt{29}},\ \frac{3}{\sqrt{29}},\ \frac{4}{\sqrt{29}}$

Hint:

The sum of squares of direction cosines must always equal exactly 1 ($\ell^2 + m^2 + n^2 = 1$).
Let's verify option (A): $\left(\frac{1}{\sqrt{14}}\right)^2 + \left(\frac{2}{\sqrt{14}}\right)^2 + \left(\frac{3}{\sqrt{14}}\right)^2 = \frac{1}{14} + \frac{4}{14} + \frac{9}{14} = \frac{14}{14} = 1$. This mathematical property lets you filter out invalid combinations instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem provides the coordinates of the origin $O(0,0,0)$ and a point $P(1,2,3)$. We need to determine the direction cosines of the vector line segment $\overline{OP}$.

Step 2: Key Formula or Approach:
For a line segment connecting the origin to a point $P(x,y,z)$, its direction ratios are simply $(x, y, z)$.
The direction cosines $(\ell, m, n)$ are computed by normalizing these direction ratios by the magnitude distance $r = \sqrt{x^2 + y^2 + z^2}$:
$$\ell = \frac{x}{r}, \quad m = \frac{y}{r}, \quad n = \frac{z}{r}$$

Step 3: Detailed Explanation:
The given coordinates are $x = 1$, $y = 2$, and $z = 3$.
First, calculate the vector distance magnitude $r$:
$$r = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$ Now, calculate each direction cosine component individually:
$$\ell = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}$$

Step 4: Final Answer:
The direction cosines of $\overline{OP}$ are $\frac{1}{\sqrt{14}},\ \frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}}$, which corresponds to option (A).