Vector Algebra
Unit Vector and Dot Product
Grade 12
Question:
<p>Given that \(\vec{u} = \hat{i} + \hat{j}\), \(\vec{v} = \hat{i} - \hat{j}\), \(\vec{\omega} = \hat{i} + 2\hat{j} + 3\hat{k}\). Let \(\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}\) be a unit vector such that \(\vec{u} \cdot \vec{n} = 0\) and \(\vec{v} \cdot \vec{n} = 0\). Find \(|\vec{\omega} \cdot \vec{n}|\).</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>0</p>
Step-by-Step Solution
Key Concept: The vector $\vec{n}$ perpendicular to both $\vec{u}$ and $\vec{v}$ must be parallel to their cross product $\vec{u} \times \vec{v}$. Since $\vec{n}$ is a unit vector, normalize this cross product to find it, then compute the dot product with $\vec{\omega}$.
Step 1: Find the direction perpendicular to both $\vec{u}$ and $\vec{v}$ using cross product. $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-1-1) = -2\hat{k}$ Step 2: Verify orthogonality conditions from the cross product result. Since $\vec{n}$ is perpendicular to both $\vec{u}$ and $\vec{v}$, it must be parallel to $\vec{u} \times \vec{v} = -2\hat{k}$. Thus $\vec{n} = \lambda\hat{k}$ for some scalar $\lambda$. Step 3: Apply the unit vector condition. $|\vec{n}| = 1 \Rightarrow |\lambda| = 1 \Rightarrow \lambda = \pm 1$ Therefore $\vec{n} = \pm\hat{k}$ Step 4: Compute $|\vec{\omega} \cdot \vec{n}|$. $\vec{\omega} \cdot \vec{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (\pm\hat{k}) = \pm 3$ $|\vec{\omega} \cdot \vec{n}| = 3$ ∴ Answer: C (which equals 3)
Correct Answer: C