Vector Algebra
Cross Product
Grade 12
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>
<li>\(\lambda=0,\;|\vec{n}|=|\vec{K}|\)</li>
<li>\(\lambda=1,\;|\vec{n}|=|\vec{K}-\vec{J}|\)</li>
<li>\(\lambda=\dfrac{\vec{J}\cdot\vec{K}}{|\vec{J}|^2}\) and \(|\vec{n}|^2=|\vec{K}|^2-\lambda^2|\vec{J}|^2\)</li>
<li>\(\lambda=\dfrac{\vec{J}\cdot\vec{K}}{|\vec{J}|^2}\)</li>
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