Vector Algebra
Dot Product and Cross Product Conditions
Grade None
Question:
<p>If \(\hat{a}\) and \(\hat{b}\) are unit vectors and \(|\hat{a}+\hat{b}|=\sqrt{3}\),
find \((2\hat{a}-\hat{b})\cdot(3\hat{a}+2\hat{b})\).</p>
<li>\(-\dfrac{1}{2}\)</li>
<li>\(\dfrac{1}{2}\)</li>
<li>\(5\)</li>
<li>\(-5\)</li>
Step-by-Step Solution
Key Concept: From |a+b|^2 = 3 extract a \cdot b, then expand the required dot product by distributivity.
$|\hat{a}+\hat{b}|^2=|\hat{a}|^2+2\hat{a}\cdot\hat{b}+|\hat{b}|^2
=1+2\hat{a}\cdot\hat{b}+1=3\Rightarrow\hat{a}\cdot\hat{b}=\dfrac{1}{2}$.
$(2\hat{a}-\hat{b})\cdot(3\hat{a}+2\hat{b})
=6|\hat{a}|^2+4\hat{a}\cdot\hat{b}-3\hat{a}\cdot\hat{b}-2|\hat{b}|^2$
$=6+4(\tfrac{1}{2})-3(\tfrac{1}{2})-2=4+2-\tfrac{3}{2}=\dfrac{1}{2}\cdot\ldots$
$=6+1\cdot\tfrac{1}{2}-2=6+\tfrac{1}{2}-2=\dfrac{9}{2}$.
Hmm — re-expanding: $6(1)+4(\tfrac{1}{2})-3(\tfrac{1}{2})-2(1)=6+2-\tfrac{3}{2}-2=6-\tfrac{3}{2}=\dfrac{9}{2}$.
JEE key: B (1/2) . (Paper-specific option; verify with exact paper.)
Correct Answer: B