Question

If \[ (\vec a+\vec b)\perp \vec b \quad \text{and} \quad (\vec a+2\vec b)\perp \vec a, \] then

• A. \(|\vec a|=2|\vec b|\)
• B. \(|\vec a|=\sqrt2|\vec b|\)
• C. \(2|\vec a|=|\vec b|\)
• D. \(|\vec a|=|\vec b|\)

Hint:

Whenever vectors are perpendicular, immediately apply the condition: \[ \vec u\cdot\vec v=0 \] and expand using distributive properties of the dot product.

Answer 1

The Correct Option is B

Concept: Two vectors are perpendicular if and only if their dot product is zero: \[ \vec u\cdot\vec v=0 \] We use this property to form equations involving vector magnitudes.

Step 1: Use the first perpendicular condition. Given: \[ (\vec a+\vec b)\perp\vec b \] Therefore, \[ (\vec a+\vec b)\cdot\vec b=0 \] Expanding: \[ \vec a\cdot\vec b+\vec b\cdot\vec b=0 \] \[ \vec a\cdot\vec b+|\vec b|^2=0 \] Hence, \[ \vec a\cdot\vec b=-|\vec b|^2 \]

Step 2: Use the second perpendicular condition. Given: \[ (\vec a+2\vec b)\perp\vec a \] Thus, \[ (\vec a+2\vec b)\cdot\vec a=0 \] Expanding: \[ \vec a\cdot\vec a+2\vec b\cdot\vec a=0 \] \[ |\vec a|^2+2(\vec a\cdot\vec b)=0 \] Substitute: \[ \vec a\cdot\vec b=-|\vec b|^2 \] Hence, \[ |\vec a|^2+2(-|\vec b|^2)=0 \] \[ |\vec a|^2=2|\vec b|^2 \] Taking square roots: \[ |\vec a|=\sqrt2|\vec b| \] Hence, \[ \boxed{|\vec a|=\sqrt2|\vec b|} \]