# The scalar product of the vector $$\hat i + \hat j + \hat k$$ with a unit vector along the sum of vectors $$2\hat i + 4 \hat j − 5\hat k$$ and $$λ\hat i + 2 \hat j + 3\hat k$$ is equal to one. Find the value of $$λ.$$

Toolbox:
• Unit vector in the direction of $\overrightarrow a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
• If $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$,then $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$
• Magnitude of unit vector is 1.
Step 1:
Let us first determine the sum of the vectors:
$(2\hat i+4\hat j-5\hat k)+(\lambda\hat i+2\hat j+3\hat k)=(2+\lambda)\hat i+(4+2)\hat j+(-5+3)\hat k$
Let us assume the sum of these vectors as $\overrightarrow a$
Hence magnitude of $\overrightarrow a$ is $|\overrightarrow a|$
$|\overrightarrow a|=\sqrt{(2+\lambda)^2+6^2+(-2)^2}$
$\quad\;\;=\sqrt{4+4\lambda+\lambda^2+36+4}$
$\quad\;\;=\sqrt{4+4\lambda+\lambda^2+40}$
$\quad\;\;=\sqrt{\lambda^2+4\lambda+44}=r$(assume)--------(1)
Step 2:
We know unit vector is $\large\frac{\overrightarrow a}{|\overrightarrow a|}$
Hence the unit vector $\overrightarrow a=\large\frac{(2+\lambda)\hat i}{r}+\large\frac{6\hat j}{r}-\large\frac{2\hat k}{r}$
It is given that this unit vector is 1.
Therefore $\large\frac{(2+\lambda)\hat i}{r}+\large\frac{6\hat j}{r}-\large\frac{2\hat k}{r}$$=1 Step 3: The scalar product of \hat i+\hat j+\hat k and this unit vector \overrightarrow a \Rightarrow (\hat i+\hat j+\hat k)\big(\large\frac{(2+\lambda)\hat i}{r}+\large\frac{6\hat j}{r}-\large\frac{2\hat k}{r}\big) Therefore \large\frac{(2+\lambda)}{r}+\large\frac{6}{r}-\large\frac{2}{r}$$=1$
Step 4:
On simplifying we get
$\large\frac{\lambda+6}{r}$$=1$
$\Rightarrow \lambda+6=r\Rightarrow (\lambda+6)^2=r^2$
Step 5:
Substituting this in equ(1) we get
$\lambda^2+4\lambda+44=(\lambda+6)^2$
$\Rightarrow \lambda^2+4\lambda+44=\lambda^2+12\lambda+36$
$8\lambda=8$
$\lambda=1$
Hence the value of $\lambda=1$