The scalar product of the vector \(\hat{\imath}+\hat{\jmath}+\hat{k}\) with the unit vector along the sum of the vectors \(\hat{\imath}+2\hat{\jmath}-3\hat{k}\) and \(\lambda\hat{\imath}+\hat{\jmath}+2\hat{k}\) is equal to 1. The value of \(\lambda\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
The sum of the two vectors is
\(\vec{s} = (1+\lambda)\hat{\imath}+3\hat{\jmath}-\hat{k}\)
\(|\vec{s}| = \sqrt{(1+\lambda)^2+9+1} = \sqrt{(1+\lambda)^2+10}\)
The unit vector along \(\vec{s}\) is \(\displaystyle\frac{\vec{s}}{|\vec{s}|}\). Its scalar product with \(\hat{\imath}+\hat{\jmath}+\hat{k}\) is 1:
\( \displaystyle \frac{(1+\lambda)+3-1}{\sqrt{(1+\lambda)^2+10}} = 1 \)
\(\lambda+3 = \sqrt{(1+\lambda)^2+10}\)
Squaring both sides:
\(\lambda^2+6\lambda+9 = \lambda^2+2\lambda+11\)
\(4\lambda = 2\)
\( \displaystyle \lambda = \frac{1}{2} \)
Check: \(\lambda+3 = \displaystyle\frac{7}{2} \gt 0\), and \(\sqrt{\displaystyle\frac{9}{4}+10} = \sqrt{\displaystyle\frac{49}{4}} = \displaystyle\frac{7}{2}\).