JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 11)
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 $$\lambda \in \mathbb{R}$$.
(B) $$\vec{a}$$ and $$\vec{c}$$ are always parallel.
Then,
only (B) is correct
both (A) and (B) are correct
only (A) is correct
neither (A) nor (B) is correct
Explanation
$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$
$$ \Rightarrow $$ $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$
$$ \begin{aligned} & \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\ & =|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{a}) \\\\ & \Rightarrow \vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=0 \Rightarrow \vec{c} \cdot \vec{a}=0 \\\\ & |\vec{a}+\lambda \vec{c}|^{2}=|\vec{a}|^{2}+\lambda^{2}|\vec{c}|^{2}+0 \geq|\vec{a}|^{2} \end{aligned} $$
So $\mathrm{A}$ is correct.
$B$ is incorrect.
$$ \Rightarrow $$ $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$
$$ \begin{aligned} & \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\ & =|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{a}) \\\\ & \Rightarrow \vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=0 \Rightarrow \vec{c} \cdot \vec{a}=0 \\\\ & |\vec{a}+\lambda \vec{c}|^{2}=|\vec{a}|^{2}+\lambda^{2}|\vec{c}|^{2}+0 \geq|\vec{a}|^{2} \end{aligned} $$
So $\mathrm{A}$ is correct.
$B$ is incorrect.
Comments (0)
