Question

Let $\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$ and $\vec{b} \cdot \vec{c}=0$. Consider the following two statements: 

(A) $|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$ for all $\lambda \in R$ 

(B) $\vec{a}$ and $\vec{c}$ are always parallel. Then. is

• A. both (A) and (B) are correct
• B. only $( A )$ is correct
• C. only (B) is correct
• D. neither $( A )$ nor $(B)$ is correct

Answer 1

The Correct Option is B

$|\vec{a}+\vec{b}+\vec{c}{|}^2=|\vec{a}+\vec{b}-\vec{c}{|}^2$
$2\vec{a}\cdot \vec{b}+2\vec{b}\cdot \vec{c}+2\vec{c}\cdot \vec{a}=2\vec{a}\cdot \vec{b}-2\vec{b}\cdot \vec{c}-2\vec{c}\cdot \vec{a}$
$4\vec{a}\cdot \vec{c}=0$
$B$ is incorrect
$|\overrightarrow{a}+\lambda \overrightarrow{c}{|}^2\ge |\overrightarrow{a}{|}^2$
${\lambda }^2{c}^2\ge 0$
True $\forall \lambda \in R$ (A) is correct.