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>
<li>\(84\)</li>
<li>\(49\)</li>
<li>\(91\)</li>
<li>\(100\)</li>
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