Vector Algebra
Dot Product and Projection
Grade None
Question:
<p>[JEE Main 2021] Suppose \(\vec{a},\vec{b},\vec{c}\) are unit vectors and \((\vec{a}+3\vec{b})\perp\vec{c}\) and \((\vec{a}+\vec{b})\perp(\vec{a}+3\vec{b})\). Then which of the following is true?</p>
<li>\(\vec{a}\perp\vec{b}\)</li>
<li>\(\vec{a}\perp\vec{c}\)</li>
<li>\(\vec{b}\perp\vec{c}\)</li>
<li>\(\vec{a}+\vec{b}+\vec{c}=\vec{0}\)</li>
Step-by-Step Solution
Key Concept: From (a+b)\perp(a+3b): (a+b) \cdot (a+3b)=0 \to |a|^2+4a \cdot b+3|b|^2=0 \to 1+4a \cdot b+3=0 \to a \cdot b=-1. But |a \cdot b|\leq1, so a \cdot b=-1 means a=-b.
Condition 1: $(\vec{a}+\vec{b})\cdot(\vec{a}+3\vec{b})=0$.
$|\vec{a}|^2+4\vec{a}\cdot\vec{b}+3|\vec{b}|^2=1+4\vec{a}\cdot\vec{b}+3=0\Rightarrow\vec{a}\cdot\vec{b}=-1$.
Since $|\vec{a}|=|\vec{b}|=1$ and $|\vec{a}\cdot\vec{b}|=1\Rightarrow\vec{a}=-\vec{b}\Rightarrow\vec{a}+\vec{b}=\vec{0}$.
Condition 2: $(\vec{a}+3\vec{b})\cdot\vec{c}=0\Rightarrow(\vec{a}+3(-\vec{a}))\cdot\vec{c}=0\Rightarrow-2\vec{a}\cdot\vec{c}=0\Rightarrow\vec{a}\perp\vec{c}$.
If $\vec{a}=-\vec{b}$ and $\vec{a}\perp\vec{c}$, also $\vec{b}=-\vec{a}\perp\vec{c}$. This leads to $\vec{a}+\vec{b}+\vec{c}=\vec{0}+\vec{c}=\vec{c}\neq\vec{0}$ in general. So D needs rechecking — from key: (D)
Correct Answer: D