Vectors
Scalar triple product with cross product conditions
MJAT_TS8_P1
Grade 12
Question:
Given $\vec{a}=2\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$, $[\vec{a}\vec{b}\vec{c}]=0$, $|\vec{c}|=2$, $\vec{c}\cdot\vec{a}=0$. If $\vec{d}$ is such that $(\vec{a}\times\vec{b})\times\vec{d}=\vec{0}$ and $|\vec{d}|=3$, then $[\vec{b}\vec{c}\vec{d}]$ equals:
Step-by-Step Solution
Key Concept: $(\vec{a}\times\vec{b})\times\vec{d}=\vec{0}\Rightarrow\vec{d}\parallel(\vec{a}\times\vec{b})$. $\vec{a}\times\vec{b}=\det[\hat{i},\hat{j},\hat{k};2,1,1;1,2,-1]=(-3,3,3)\propto(-1,1,1)$. $|\vec{d}|=3$, so $\vec{d}=\pm(-1,1,1)$.
$[\vec{b}\vec{c}\vec{d}]=\mathbf{3}$.
Correct Answer: 3