Question

Let \( \vec{a} = \hat{i} + x\hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \) and \( |\vec{a}+\vec{b}| = |\vec{a}| + |\vec{b}| \) then

• A. \( x=1 \)
• B. \( x=-1 \)
• C. \( x=0 \)
• D. No such real x exits (Telugu: ఇలాంటి వాస్తవసంఖ్య x వ్యవస్థితం కాదు)

Hint:

The triangle inequality for vectors states \(||\vec{u}+\vec{v}|| \le ||\vec{u}|| + ||\vec{v}||\).
Equality holds, \(||\vec{u}+\vec{v}|| = ||\vec{u}|| + ||\vec{v}||\), if and only if \(\vec{u}\) and \(\vec{v}\) are in the same direction (i.e., one is a non-negative scalar multiple of the other, or one is the zero vector).
If \(\vec{a} = k\vec{b}\) with \(k \ge 0\), then \(|\vec{a}+\vec{b}| = |(k+1)\vec{b}| = (k+1)|\vec{b}|\) (since \(k+1>0\)). And \(|\vec{a}|+|\vec{b}| = |k\vec{b}|+|\vec{b}| = k|\vec{b}|+|\vec{b}| = (k+1)|\vec{b}|\) (since \(k \ge 0\)).

Answer 1

The Correct Option is B

Condition: The equality \[ |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \] holds if and only if vectors \( \vec{a} \) and \( \vec{b} \) are collinear and in the same direction — that is, \( \vec{a} = k\vec{b} \) for some scalar \( k \ge 0 \), or \( \vec{b} = m\vec{a} \) for \( m \ge 0 \), or one of the vectors is a zero vector.

Given:

\[ \vec{a} = \hat{i} + x\hat{j} + \hat{k}, \quad \vec{b} = \hat{i} - \hat{j} + \hat{k} \] Since \( \vec{b} \neq \vec{0} \), for collinearity in the same direction we assume: \[ \vec{a} = k\vec{b} \text{ for some } k > 0 \] Substituting: \[ \hat{i} + x\hat{j} + \hat{k} = k(\hat{i} - \hat{j} + \hat{k}) = k\hat{i} - k\hat{j} + k\hat{k} \]

Comparing components:

• \( \hat{i} \): \( 1 = k \Rightarrow k = 1 \)
• \( \hat{k} \): \( 1 = k \Rightarrow k = 1 \) — consistent
• \( \hat{j} \): \( x = -k = -1 \)

So, the value of \( x \) is:

\[ \boxed{x = -1} \]

Verification:

If \( x = -1 \), then \[ \vec{a} = \hat{i} - \hat{j} + \hat{k} = \vec{b} \]

Now compute magnitudes:

\[ |\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}, \quad |\vec{b}| = \sqrt{3} \] \[ \vec{a} + \vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k} \Rightarrow |\vec{a} + \vec{b}| = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{12} = 2\sqrt{3} \] \[ |\vec{a}| + |\vec{b}| = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \Rightarrow |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \]

Hence, verified. The correct value is:

\( \boxed{x = -1} \)