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

Hint:

When dealing with vector magnitudes and dot products, remember that the square of the magnitude of a vector is always non-negative. Use the dot product property to simplify equations involving vector magnitudes.

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.

Answer 2

Step 1: Start with the given equation: \[ \left| \mathbf{a} + \mathbf{b} + \mathbf{c} \right| = \left| \mathbf{a} + \mathbf{b} - \mathbf{c} \right|. \] Square both sides of the equation to eliminate the absolute value: \[ \left( \mathbf{a} + \mathbf{b} + \mathbf{c} \right)^2 = \left( \mathbf{a} + \mathbf{b} - \mathbf{c} \right)^2. \] Now expand both sides: \[ \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} + 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 + 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2 = \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 - 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2. \] Simplify the equation by canceling out the common terms: \[ 2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} = -2 \mathbf{a} \cdot \mathbf{c} - 2 \mathbf{b} \cdot \mathbf{c}. \] This simplifies further to: \[ 4 \mathbf{a} \cdot \mathbf{c} = 0 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{c} = 0. \] This shows that \( \mathbf{a} \) and \( \mathbf{c} \) are perpendicular to each other (i.e., \( \mathbf{a} \cdot \mathbf{c} = 0 \)). 
Step 2: Therefore, \( \mathbf{a} \) and \( \mathbf{c} \) are perpendicular, not parallel. Hence, statement (B) is incorrect. 
Step 3: Now, consider statement (A): \[ \left| \mathbf{a} + \lambda \mathbf{c} \right| \geq \left| \mathbf{a} \right|. \] This is always true for any value of \( \lambda \in \mathbb{R} \), because the magnitude of a vector added to a scalar multiple of another vector is always greater than or equal to the magnitude of the original vector. This follows from the triangle inequality, which states that the length of the sum of two vectors is at least as large as the length of either vector individually. Therefore, statement (A) is correct.