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.

Step 1 — The governing identity

If a line makes angles $\alpha, \beta, \gamma$ with the $x$-, $y$- and $z$-axes, its direction cosines are $l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma$, and they always satisfy

(This is just the statement that the unit vector along the line has magnitude $1$.)

Step 2 — Substitute the given angles

Step 3 — Solve for $\theta$

Step 4 — Apply the acuteness condition

$\theta$ is given to be acute, so $\cos\theta > 0$:

(The rejected root $\cos\theta = -\tfrac{\sqrt3}{2}$ would give the obtuse $\theta = 150^\circ$, which the question excludes.)

ANSWER: A

The key idea: The cosine of the angle between any two vectors is given by their dot product divided by the product of their magnitudes. Here, one vector is $\overrightarrow{PQ}$ and the other is the direction of the y‑axis, which is simply the unit vector $\hat{\jmath}$. So we just need the y‑component of the unit vector along $\overrightarrow{PQ}$.

1. Find $\overrightarrow{PQ}$

   $\overrightarrow{PQ} = \text{position of } Q - \text{position of } P$

   $= (5\hat{\imath} -2\hat{\jmath} +4\hat{k}) - (\hat{\imath} +3\hat{\jmath} -7\hat{k})$

   $= (5-1)\hat{\imath} + (-2-3)\hat{\jmath} + (4+7)\hat{k}$

   $= 4\hat{\imath} -5\hat{\jmath} +11\hat{k}$.

2. Magnitude of $\overrightarrow{PQ}$

   $|\overrightarrow{PQ}| = \sqrt{4^2 + (-5)^2 + 11^2} = \sqrt{16 + 25 + 121} = \sqrt{162}$.

3. Cosine of the angle with the y‑axis

   The y‑axis direction vector is $\hat{\jmath}$. The cosine formula:

   $\cos\theta = \frac{\overrightarrow{PQ} \cdot \hat{\jmath}}{|\overrightarrow{PQ}| \cdot |\hat{\jmath}|}$.

   Since $|\hat{\jmath}| = 1$ and $\overrightarrow{PQ} \cdot \hat{\jmath} = -5$, we get

   $\cos\theta = \frac{-5}{\sqrt{162}}$.

4. Interpretation

   The negative sign means the angle is obtuse — $\overrightarrow{PQ}$ points generally downward along the y‑direction relative to the positive y‑axis.

We are told the line passes through the origin and has direction angles $\frac{2\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3}$.

The direction cosines are the cosines of these angles, and they tell us the components of a unit vector along the line.

The symmetric form of a line through the origin is $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$, where $(a,b,c)$ are any direction ratios proportional to the direction cosines.

1. Compute the direction cosines

So the direction cosines are $\left(-\frac12,\ \frac{\sqrt2}{2},\ \frac12\right)$.

2. Convert to direction ratios We can multiply by 2 to clear denominators:

Any scalar multiple works; this set is simplest.

3. Write the symmetric equation Through the origin:

Equivalently,

But the options are given in the form $x = \frac{y}{\text{something}} = \pm z$.

Multiply the first equality by $-1$:

That matches option (D) at first glance — but check carefully:

From $\frac{x}{-1} = \frac{y}{\sqrt2}$, cross-multiplying gives $x = -\frac{y}{\sqrt2}$, i.e. $x = \frac{y}{-\sqrt2}$.

From $\frac{x}{-1} = z$, we get $x = -z$.

So the full set is $x = \frac{y}{-\sqrt2} = -z$. That is exactly option (D).

4. Verify the other options
   • (A) $x = \frac{y}{\sqrt2} = z$ would mean direction ratios $(1, \sqrt2, 1)$, which gives cosines $(\frac12, \frac{\sqrt2}{2}, \frac12)$ — the first cosine is positive, but we need $-\frac12$.
   • (B) $\frac{x}{-1} = \frac{y}{-\sqrt2} = z$ gives ratios $(-1, -\sqrt2, 1)$ — the second cosine would be negative, but we need positive $\frac{\sqrt2}{2}$.
   • (C) $x = \frac{y}{-\sqrt2} = z$ gives ratios $(1, -\sqrt2, 1)$ — the second cosine negative again. Only (D) matches all three signs.

Direction-cosine identity

The line makes angle $\theta$ with both the $x$- and $z$-axes and angle $\beta$ with the $y$-axis, so its direction cosines satisfy

Therefore

Apply the given condition

With $\sin^2\beta=3\sin^2\theta=3(1-\cos^2\theta)$:

We are told the vector $\vec{r}$ has magnitude $|\vec{r}| = 3\sqrt{2}$ and makes an angle of $\frac{\pi}{4}$ with the $y$-axis and $\frac{\pi}{2}$ with the $z$-axis. The angle with the $x$-axis is not directly given, but we can find it using the fundamental relation between direction cosines.

Concept & Intuition

For any vector in 3D, the cosines of the angles it makes with the $x$, $y$, and $z$ axes are called direction cosines, often denoted $\cos\alpha$, $\cos\beta$, $\cos\gamma$. These satisfy the identity:

This is because the components are $|\vec{r}|\cos\alpha$, $|\vec{r}|\cos\beta$, $|\vec{r}|\cos\gamma$, and the sum of their squares equals $|\vec{r}|^2$. Once we know two angles, we can solve for the third — but note the sign ambiguity: the cosine could be positive or negative, giving two possible directions.

1. Write down what we know

   Angle with $y$-axis: $\beta = \frac{\pi}{4}$, so $\cos\beta = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

   Angle with $z$-axis: $\gamma = \frac{\pi}{2}$, so $\cos\gamma = \cos\frac{\pi}{2} = 0$.

2. Use the direction cosine identity

Substitute the known values:

Hence $\cos\alpha = \pm \frac{1}{\sqrt{2}}$.

3. Find the components The vector components are:

4. Write the vector So $\vec{r} = \pm 3\hat{\imath} + 3\hat{\jmath} + 0\hat{k}$, which simplifies to $\pm 3\hat{\imath} + 3\hat{\jmath}$.

5. Match with the options The vector $\pm 3\hat{\imath} + 3\hat{\jmath}$ appears exactly as option (D).

$E$ is the midpoint of $AC$ with $A(-2,3,6),\,C(0,5,8)$:

Then with $B(-4,4,9)$,

Direction cosines: