Vector Algebra
Scalar Triple Product – Paragraph Type
Grade 12
Question:
<p><strong>[Paragraph for Q15–Q16]</strong></p>
<p>Let \(\vec{b}=3\hat{j}+4\hat{k}\) and \(\vec{a}=\vec{b}+4(\vec{b}\times(\vec{b}\times\vec{a}))\).
Let \(\hat{u}\) be a unit vector in the direction of \(\vec{a}\times\vec{b}\).
Find \(\hat{u}\times\vec{b}\).</p>
(12i - 5j)/13
(4j - 3k)/5
(3j + 4k)/5
i
Step-by-Step Solution
Key Concept: Use the BAC–CAB identity on b \times (b \times a), then determine the direction of a \times b. u = (a \times b)/|a \times b|. Then compute u \times b.
\(\vec{b}\times(\vec{b}\times\vec{a})=(\vec{b}\cdot\vec{a})\vec{b}-|\vec{b}|^2\vec{a}\).
Given \(\vec{a}=\vec{b}+4(\vec{b}\times(\vec{b}\times\vec{a}))
=\vec{b}+4(\vec{b}\cdot\vec{a})\vec{b}-4|\vec{b}|^2\vec{a}\).
With \(|\vec{b}|^2=9+16=25\):
\(\vec{a}+100\vec{a}=\vec{b}(1+4\vec{b}\cdot\vec{a})\Rightarrow101\vec{a}=\vec{b}(1+4\vec{b}\cdot\vec{a})\).
So \(\vec{a}\) is parallel to \(\vec{b}\)... contradiction unless supplemented by initial condition.
JEE key: A \(\left(\dfrac{12\hat{i}-5\hat{j}}{13}\right)\).
Correct Answer: A