Question

If $|\vec{u}| = 2$ and $\vec{u}$ makes angles of $60^\circ$ and $120^\circ$ with the axes $OX$ and $OY$ at the origin, then $\vec{u} =$

• A. $\hat{i} + \hat{j} + 2\hat{k}$
• B. $2(\hat{i} + \hat{j} \pm \sqrt{2}\hat{k})$
• C. $2(\hat{i} - \hat{j} + \sqrt{2}\hat{k})$
• D. $\hat{i} - \hat{j} \pm \sqrt{2}\hat{k}$

Hint:

You can use option elimination to find the answer without calculating the third angle! Compute the first two components directly: $u_x = |\vec{u}|\cos(60^\circ) = 2(0.5) = 1$, and $u_y = |\vec{u}|\cos(120^\circ) = 2(-0.5) = -1$. Looking at the choices, only option (D) features $+1\hat{i} - 1\hat{j}$ at the front, saving you valuable time during the test!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the overall magnitude of a 3D vector $\vec{u}$ alongside its direction angles with the positive $X$-axis ($\alpha = 60^\circ$) and $Y$-axis ($\beta = 120^\circ$). We must determine the vector expression by computing its direction cosines and components.

Step 2: Key Formula or Approach:
1. Let the direction cosines of the vector be $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$. 2. Use the fundamental direction cosine identity: $$l^2 + m^2 + n^2 = 1$$ 3. Once $l$, $m$, and $n$ are known, the vector can be reconstructed as: $$\vec{u} = |\vec{u}|(l\hat{i} + m\hat{j} + n\hat{k})$$

Step 3: Detailed Explanation:
Let's compute the known direction cosines first: $$l = \cos(60^\circ) = \frac{1}{2}$$ $$m = \cos(120^\circ) = -\frac{1}{2}$$ Now, substitute these into the sum-of-squares identity to solve for $n$: $$\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + n^2 = 1$$ $$\frac{1}{4} + \frac{1}{4} + n^2 = 1$$ $$\frac{1}{2} + n^2 = 1 \implies n^2 = \frac{1}{2} \implies n = \pm\frac{1}{\sqrt{2}}$$ With the magnitude $|\vec{u}| = 2$, let's build the vector components: $$\vec{u} = 2 \left( \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} \pm \frac{1}{\sqrt{2}}\hat{k} \right)$$ Distributing the scalar factor of 2 inside the brackets: $$\vec{u} = \hat{i} - \hat{j} \pm \frac{2}{\sqrt{2}}\hat{k} = \hat{i} - \hat{j} \pm \sqrt{2}\hat{k}$$

Step 4: Final Answer:
The full component form of the vector is $\hat{i} - \hat{j} \pm \sqrt{2}\hat{k}$, which corresponds to option (D).