# The scalar product of the vector $\overrightarrow a = \hat i + \hat j + \hat k$ with a unit vector along the sums of vectors $\overrightarrow b = 2 \hat i + 4\hat j - 5 \hat k$ and $\overrightarrow c = \lambda \hat i + 2 \hat j + 3 \hat k\;$ is equal to one. Find the value of $\lambda$ and hence find the unit vector along $\overrightarrow b + \overrightarrow c$

$\begin{array}{1 1} \frac{3 \hat i}{7} +\frac{6 \hat j}{7} -\frac{2 \hat k}{7} \\ \frac{4 \hat i}{7} +\frac{6 \hat j}{7} -\frac{2 \hat k}{7} \\ \frac{2 \hat i}{7} +\frac{6 \hat j}{7} -\frac{ \hat k} {7} \\ \frac{3 \hat i}{7} -\frac{3 \hat j}{7} -\frac{2 \hat k}{7}\end{array}$

Solution :
$\overrightarrow a = \hat i + \hat j + \hat k$
$\overrightarrow b = 2 \hat i + 4\hat j - 5 \hat k$
$\overrightarrow c = \lambda \hat i + 2 \hat j + 3 \hat k$
$\overrightarrow b + \overrightarrow c$$= (2+\lambda) \hat i + 6 \hat j - 2 \hat k with vector along \overrightarrow b + \overrightarrow c = \large\frac{(2 +\lambda) \hat i + 5 \hat j - 2 \hat k}{\sqrt{(2+\lambda)^2+36+4}} \qquad= \large\frac{(2+ \lambda ) \hat i +6 \hat j - 2 \hat k}{\sqrt{ (2+\lambda)^2+40}} \hat i+ \hat j +\hat k = \large\frac{(2+ \lambda) \hat i +6 \hat j - 2 \hat k}{\sqrt {(2+\lambda)^2+40}}$$=1$
$\large\frac{(2+\lambda)+6-2}{\sqrt{(2+\lambda)^2+40}}$$=1 \qquad= (2+ \lambda)+4 =\sqrt{(2+ \lambda)^2+40} 6+ \lambda = \sqrt{(2+\lambda)^2+40} 36+ \lambda^2+12 \lambda = (2 + \lambda)^2+40 36+ \lambda^2+12 \lambda = 4 + \lambda^2+4 \lambda +40 8 \lambda= 8 \lambda = \large\frac{8}{8}$$=1$
$\lambda =1$
Unit vector =$\large\frac{(2+ 1) \hat i +6 \hat j - 2 \hat k}{\sqrt{ (2+1)^2+40}}$
$\qquad = \large\frac{3 \hat i}{7} +\frac{6 \hat j}{7} - \frac{2 \hat k}{7}$