Vector Algebra
Cross Product
Grade None
Question:
<p>Suppose \(\vec{J}=\hat{i}-2\hat{j}+\hat{k}\) and \(\vec{K}=3\hat{i}+\hat{j}-\hat{k}\). If \(\vec{K}=\lambda\vec{J}+\vec{n}\) where \(\vec{n}\perp\vec{J}\), then \(\lambda\) and \(|\vec{n}|\) equal</p>
\(\lambda=0,\;|\vec{n}|=|\vec{K}|\)
\(\lambda=1,\;|\vec{n}|=|\vec{K}-\vec{J}|\)
\(\lambda=\dfrac{\vec{J}\cdot\vec{K}}{|\vec{J}|^2}\) and \(|\vec{n}|^2=|\vec{K}|^2-\lambda^2|\vec{J}|^2\)
\(\lambda=\dfrac{\vec{J}\cdot\vec{K}}{|\vec{J}|^2}\)
Step-by-Step Solution
Key Concept: K=\lambdaJ+n with n\perpJ. Dot both sides with J: J \cdot K=\lambda|J|^2, so \lambda=J \cdot K/|J|^2. Then n=K-\lambdaJ and |n|^2=|K|^2-\lambda^2|J|^2.
Given $\vec{K}=\lambda\vec{J}+\vec{n}$ with $\vec{n}\cdot\vec{J}=0$.
Dot with $\vec{J}$: $\vec{J}\cdot\vec{K}=\lambda|\vec{J}|^2\Rightarrow\lambda=\dfrac{\vec{J}\cdot\vec{K}}{|\vec{J}|^2}$. ✓ (C, D)
Compute: $\vec{J}\cdot\vec{K}=3-2-1=0\Rightarrow\lambda=0$. ✓ (A: $\lambda=0,\;|\vec{n}|=|\vec{K}|$)
Options A, C, D are correct. Answer: ACD
Correct Answer: ACD