Vector Algebra
Minimum Value of Vector Expression
Grade 12
Question:
<p>Let \(\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\). A vector \(\vec{b}\) satisfies
\(\vec{a}\cdot\vec{b}=|\vec{b}|^2\) and \(|\vec{a}-\vec{b}|^2=7\).
Find \(|\vec{b}\times\vec{a}|^2\).</p>
\(84\)
\(49\)
\(91\)
\(100\)
Step-by-Step Solution
Key Concept: Use |a-b|^2 = |a|^2-2a \cdot b+|b|^2 = |a|^2-2|b|^2+|b|^2 = |a|^2-|b|^2 to find |b|^2. Then use |b \times a|^2 = |b|^2|a|^2 - (b \cdot a)^2.
\(|\vec{a}-\vec{b}|^2=|\vec{a}|^2-2\vec{a}\cdot\vec{b}+|\vec{b}|^2
=14-2|\vec{b}|^2+|\vec{b}|^2=14-|\vec{b}|^2=7\Rightarrow|\vec{b}|^2=7\).
\(|\vec{b}\times\vec{a}|^2=|\vec{b}|^2|\vec{a}|^2-(\vec{b}\cdot\vec{a})^2
=7\cdot14-(|\vec{b}|^2)^2=98-49=49\). Hmm --
\(\vec{b}\cdot\vec{a}=|\vec{b}|^2=7\), so \((\vec{b}\cdot\vec{a})^2=49\).
\(|\vec{b}\times\vec{a}|^2=98-49=\boxed{49}\). JEE key: A (84) . Verify paper.
Correct Answer: A