Let \(\vec{a}=2\hat{\imath}-\hat{\jmath}+3\hat{k}\), \(\vec{b}=\hat{\imath}+\hat{\jmath}-\hat{k}\) and \(\vec{c}=2\hat{\imath}+\hat{\jmath}+\hat{k}\). If the vector \(\vec{a}+\lambda\vec{b}\) is perpendicular to \(\vec{c}\), then the value of \(\lambda\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −3
The vector \(\vec{a}+\lambda\vec{b}\) is
\(\vec{a}+\lambda\vec{b} = (2+\lambda)\hat{\imath}+(\lambda-1)\hat{\jmath}+(3-\lambda)\hat{k}\)
It is perpendicular to \(\vec{c}\), so the scalar product is zero:
\(2(2+\lambda)+1(\lambda-1)+1(3-\lambda) = 0\)
\(4+2\lambda+\lambda-1+3-\lambda = 0\)
\(2\lambda+6 = 0\)
\(\lambda = -3\)