Vector Algebra Question 15
Question 15 - 31 January - Shift 1
Let $\overrightarrow{{}a}=2 \hat{i}+\hat{j}+\hat{k}$, and $\overrightarrow{{}b}$ and $\overrightarrow{{}c}$ be two nonzero vectors such that $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$ and $\overrightarrow{{}b} \cdot \overrightarrow{{}c}=0$. Consider the following two statement:
(A) $|\overrightarrow{{}a}+\lambda \overrightarrow{{}c}| \geq|\overrightarrow{{}a}|$ for all $\lambda \in \mathbb{R}$.
(B) $\vec{a}$ and $\vec{c}$ are always parallel
(1) only (B) is correct
(2) neither (A) nor (B) is correct
(3) only (A) is correct
(4) both (A) and (B) are correct.
Show Answer
Answer: (3)
Solution:
Formula: Properties of Scalar product of vectors
$|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$
$2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}=2 \vec{a} \cdot \vec{b}-2 \vec{b} \cdot \vec{c}-2 \vec{c} \cdot \vec{a}$
$4 \vec{a} . \vec{c}=0$
$B$ is incorrect
$|\overrightarrow{{}a}+\lambda \overrightarrow{{}c}|^{2} \geq|\overrightarrow{{}a}|^{2}$
$\lambda^{2} c^{2} \geq 0$
True $\forall \lambda \in \mathbb{R}$
(A) is correct.
