Vector Algebra
Triple / Iterated Cross Products
nta_pyq_2024_jan
Grade 12
Question:
Let $\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$. Let $\vec{c}=\left(\left(\left(\vec{a}\times\vec{b}\right)\times\hat{i}\right)\times\hat{i}\right)\times\hat{i}$. Then $\vec{c}\cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to:
Step-by-Step Solution
Key Concept: Use the vector triple product identity $\vec{u}\times\hat{i}$ repeatedly. First: $(\vec{a}\times\vec{b})\times\hat{i}=(\vec{a}\cdot\hat{i})\vec{b}-(\vec{b}\cdot\hat{i})\vec{a}=-5\vec{b}-\vec{a}$. Then apply $\times\hat{i}$ two more times using $(p\hat{j}+q\hat{k})\times\hat{i}=-p\hat{k}+q\hat{j}$ and again.
$(\vec{a}\times\vec{b})\times\hat{i}=(\vec{a}\cdot\hat{i})\vec{b}-(\vec{b}\cdot\hat{i})\vec{a}=-5\vec{b}-\vec{a}=-11\hat{j}+23\hat{k}$.
$(-11\hat{j}+23\hat{k})\times\hat{i}=11\hat{k}+23\hat{j}$.
$(11\hat{k}+23\hat{j})\times\hat{i}=11\hat{j}-23\hat{k}=\vec{c}$.
$\vec{c}\cdot(-\hat{i}+\hat{j}+\hat{k})=0+11-23=-12$.
Correct Answer: 1