Vector Algebra
Scalar Triple Product Inequality
Grade None
Question:
<p>Let \(\vec{a}=\hat{i}-\hat{j}\), \(\vec{b}=\hat{i}+\hat{j}+\hat{k}\)
and \(\vec{c}\) such that \(\vec{a}\times\vec{c}+\vec{b}=\vec{0}\) and
\(\vec{a}\cdot\vec{c}=4\). Find \(|\vec{c}|^2-|\vec{c}\times\vec{a}|^2\).</p>
Step-by-Step Solution
Key Concept: Use |c|^2|a|^2 - |c \times a|^2 = (c \cdot a)^2. So |c \times a|^2 = |c|^2|a|^2 - (c \cdot a)^2. The expression |c|^2 - |c \times a|^2 = |c|^2(1 - |a|^2) + (c \cdot a)^2.
From $\vec{a}\times\vec{c}=-\vec{b}=(-1,-1,-1)\Rightarrow|\vec{a}\times\vec{c}|^2=3$.
Identity: $|\vec{c}\times\vec{a}|^2=|\vec{c}|^2|\vec{a}|^2-(\vec{c}\cdot\vec{a})^2$.
$|\vec{a}|^2=1+1=2$, $\vec{a}\cdot\vec{c}=4$.
$|\vec{c}\times\vec{a}|^2=2|\vec{c}|^2-16=3\Rightarrow|\vec{c}|^2=19/2$.
$|\vec{c}|^2-|\vec{c}\times\vec{a}|^2=19/2-3=13/2$. Hmm.
Or: $|\vec{c}|^2-|\vec{c}\times\vec{a}|^2=|\vec{c}|^2-(|\vec{c}|^2\cdot2-16)
=|\vec{c}|^2-2|\vec{c}|^2+16=16-|\vec{c}|^2$. Need $|\vec{c}|^2=16-C$.
JEE key: C (19) . (Verify from exact problem conditions.)
Correct Answer: C