Question

If \[ \vec{a}=x\hat{i}+2\hat{j}-\hat{k}\quad (x>0) \] and \[ \vec{b}=2\hat{i}-\hat{j}+2\hat{k} \] are two vectors such that \[ |\vec{a}-2\vec{b}|=|2\vec{a}+\vec{b}|, \] then \(x=\)

• A. \(4\)
• B. \(1\)
• C. \(3\)
• D. \(2\) \bigskip

Hint:

Whenever magnitudes of vectors are equal, square both sides first: \[ |\vec{u}|=|\vec{v}| \Rightarrow |\vec{u}|^2=|\vec{v}|^2. \] This converts the problem into a simple algebraic equation without square roots.

Answer 1

The Correct Option is D

Concept: If two vectors have equal magnitudes, then the squares of their magnitudes are also equal. That is, \[ |\vec{u}|=|\vec{v}| \] implies \[ |\vec{u}|^2=|\vec{v}|^2. \] This avoids square roots and simplifies calculations.

Step 1: Find \(\vec{a}-2\vec{b}\). Given \[ \vec{a}=x\hat{i}+2\hat{j}-\hat{k}, \] \[ \vec{b}=2\hat{i}-\hat{j}+2\hat{k}. \] Therefore, \[ \vec{a}-2\vec{b} = (x\hat{i}+2\hat{j}-\hat{k}) -2(2\hat{i}-\hat{j}+2\hat{k}). \] \[ = x\hat{i}+2\hat{j}-\hat{k} -4\hat{i}+2\hat{j}-4\hat{k}. \] \[ = (x-4)\hat{i}+4\hat{j}-5\hat{k}. \] Hence, \[ |\vec{a}-2\vec{b}|^2 = (x-4)^2+4^2+(-5)^2. \] \[ = (x-4)^2+41. \]

Step 2: Find \(2\vec{a}+\vec{b}\). \[ 2\vec{a}+\vec{b} = 2(x\hat{i}+2\hat{j}-\hat{k}) +(2\hat{i}-\hat{j}+2\hat{k}). \] \[ = 2x\hat{i}+4\hat{j}-2\hat{k} +2\hat{i}-\hat{j}+2\hat{k}. \] \[ = (2x+2)\hat{i}+3\hat{j}. \] Thus, \[ |2\vec{a}+\vec{b}|^2 = (2x+2)^2+3^2. \] \[ = (2x+2)^2+9. \]

Step 3: Use the given condition. Since \[ |\vec{a}-2\vec{b}|=|2\vec{a}+\vec{b}|, \] we have \[ (x-4)^2+41 = (2x+2)^2+9. \] \[ x^2-8x+16+41 = 4x^2+8x+4+9. \] \[ x^2-8x+57 = 4x^2+8x+13. \] \[ 3x^2+16x-44=0. \]

Step 4: Solve the quadratic equation. \[ 3x^2+16x-44 = (3x+22)(x-2). \] \[ (3x+22)(x-2)=0. \] Therefore, \[ x=-\frac{22}{3} \quad \text{or} \quad x=2. \] Since \[ x>0, \] we get \[ x=2. \]

Step 5: Write the final answer. \[ \boxed{2} \]