Q.Find the position vector of a point $A$ in space such that $\vec{OA}$ is inclined at $60^\circ$ to $OX$ and at $45^\circ$ to $OY$ and $|\vec{OA}| = 10$ units.

Concept: Direction Vectors — the direction cosines of $\vec{OA}$ are determined by the given angles with the axes.

Step 1: Let $\vec{OA} = 10(\cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k})$, where $\alpha, \beta, \gamma$ are the angles with $OX, OY, OZ$ respectively.

Given $\alpha = 60^\circ$, $\beta = 45^\circ$.

Step 2: Direction cosines satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

So $\left(\frac12\right)^2 + \left(\frac1{\sqrt2}\right)^2 + \cos^2\gamma = 1$

$\Rightarrow \frac14 + \frac12 + \cos^2\gamma = 1$

$\Rightarrow \cos^2\gamma = \frac14 \Rightarrow \cos\gamma = \pm\frac12$.

Step 3: Thus $\vec{OA} = 10\left(\frac12\hat{i} + \frac1{\sqrt2}\hat{j} \pm \frac12\hat{k}\right)$.

Why Direction Cosines Work

When a vector makes known angles with the coordinate axes, its components are simply the product of its magnitude and the cosines of those angles. This is because the cosine of the angle between a vector and an axis gives the fraction of the vector's length that lies along that axis. For a vector $\vec{r}$ of length $r$ making angles $\alpha, \beta, \gamma$ with the $X, Y, Z$ axes respectively:

The numbers $\cos\alpha, \cos\beta, \cos\gamma$ are called direction cosines, and they always satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. This identity is our key constraint — it lets us find the missing angle.

Step-by-Step Solution

1. Identify what we know.

   The vector $\vec{OA}$ has magnitude $|\vec{OA}| = 10$. It makes $60^\circ$ with $OX$ and $45^\circ$ with $OY$. The angle with $OZ$ is not given — we must find it.

2. Write the direction cosines for the known angles.

3. Use the fundamental identity to find the third direction cosine. Let $\gamma$ be the angle with $OZ$. Then:

4. Assemble the components. Multiply each direction cosine by the magnitude $10$:

5. Simplify the $Y$-component.

So the final expression is:

Method: Recovering a vector from two axis angles and its length

Use this when a vector's angles with two axes and its magnitude are given, and you must reconstruct the vector.

Steps

Step 1: Write the known direction cosines.

$l=\cos\alpha$ and $m=\cos\beta$ for the two given axis angles.

Step 2: Find the missing direction cosine from the identity.

Direction cosines satisfy

so $n^2=1-l^2-m^2$ and $n=\pm\sqrt{1-l^2-m^2}$. Keep BOTH signs unless the problem fixes the third angle — they give two valid vectors.

Step 3: Scale by the magnitude.

The vector is

Simplify surds (e.g. $\tfrac{10}{\sqrt2}=5\sqrt2$) and confirm $|\vec{v}|$ by recomputing the magnitude of your components.

Common Mistakes

Mistake 1: Keeping only the positive root for the third direction cosine.

Why it's wrong: $\cos^2\gamma=\frac14$ gives $\cos\gamma=\pm\frac12$, and since the inclination to $OZ$ is unspecified, BOTH signs are valid. Correct approach: keep the $\pm$, giving a $z$-component of $\pm5$.

Mistake 2: Forgetting to multiply the direction cosines by the magnitude.

Why it's wrong: $(l,m,n)$ is only a unit direction; the actual vector is $10(l,m,n)$. Correct approach: $\vec{OA}=10\left(\frac12,\frac{1}{\sqrt2},\pm\frac12\right)$.

Mistake 3: Leaving the $y$-component as $\frac{10}{\sqrt2}$.

Why it's wrong: it is not fully simplified. Correct approach: rationalise to $5\sqrt2$, giving $\vec{OA}=5\hat{i}+5\sqrt2\hat{j}\pm5\hat{k}$.