If \(\vec{a}=2\hat{\imath}-\hat{\jmath}+3\hat{k}\) and \(\vec{b}=\lambda\hat{\imath}+4\hat{\jmath}+2\hat{k}\) are such that \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\), then the value of \(\lambda\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −1
Squaring both sides of the given condition:
\(|\vec{a}+\vec{b}|^2 = |\vec{a}-\vec{b}|^2\)
\(|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b} = |\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b}\)
\(4\,\vec{a}\cdot\vec{b} = 0\)
So \(\vec{a}\) and \(\vec{b}\) are perpendicular, and their scalar product is zero:
\((2)(\lambda)+(-1)(4)+(3)(2) = 0\)
\(2\lambda+2 = 0\)
\(\lambda = -1\)