Question

Let \[ \vec{a}=\hat{i}+x\hat{j}+\hat{k}, \qquad \vec{b}=\hat{i}+\hat{j}+\hat{k} \] and \[ |\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|. \] Then

• A. \(x=1\)
• B. \(x=-1\)
• C. \(x=0\)
• D. No such real \(x\) exists

Hint:

The equality \[ |\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}| \] holds only when the vectors are parallel and point in the same direction.

Answer 1

The Correct Option is A

Step 1: Use the condition for equality in the triangle inequality.
For vectors, \[ |\vec{a}+\vec{b}|\leq |\vec{a}|+|\vec{b}|. \] Equality holds if and only if \(\vec{a}\) and \(\vec{b}\) are in the same direction, i.e., \[ \vec{a}=\lambda \vec{b} \] for some \(\lambda \ge 0\).

Step 2: Compare the components of the vectors.
Given, \[ \vec{a}=(1,x,1), \qquad \vec{b}=(1,1,1). \] If \[ \vec{a}=\lambda \vec{b}, \] then \[ (1,x,1)=(\lambda,\lambda,\lambda). \] Comparing the first and third coordinates, \[ \lambda=1. \]

Step 3: Determine \(x\).
Using \(\lambda=1\), \[ x=\lambda=1. \] Hence, \[ x=1. \]

Step 4: Verification.
For \(x=1\), \[ \vec{a}=(1,1,1)=\vec{b}. \] Therefore, \[ |\vec{a}+\vec{b}| = |2\vec{a}| = 2|\vec{a}| = |\vec{a}|+|\vec{b}|, \] which satisfies the given condition.

Step 5: Final conclusion.
Therefore, \[ \boxed{x=1} \] Hence, the correct option is \[ \boxed{(1)}. \]