Question

$2i+3j+k$ as a sum of two vectors out of which one is perpendicular to $2i-4j+k$ and another is parallel to $2i-4j+k$ is $p(8i+5j+4k)$ + $q(2i-4j+k)$ then p+q =

Answer 1

$Let$
$\overrightarrow{b}=2i-4j+k$
$\overrightarrow{a}=2i+3j+k$
$Let$
$\overrightarrow{a}={\overrightarrow{a}}_1+{\overrightarrow{a}}_2$
${\overrightarrow{a}}_1=v\text{ector parallel to}\overrightarrow{\text{b}}$
${\overrightarrow{a}}_2=v\text{ector perpendicular to}\overrightarrow{\text{b}}$
$\sin ce{\overrightarrow{a}}_{1}is\text{parallel to}\overrightarrow{\text{b}}$
${\overrightarrow{a}}_1=\lambda (2i-4j+k)$
${\overrightarrow{a}}_1=2\lambda i-4\lambda j+\lambda k$
$also$
$\overrightarrow{a_2}=\overrightarrow{a}+{\overrightarrow{a}}_1$
$\overrightarrow{a_2}=(2i+3j+k)-(2\lambda i-4\lambda j+\lambda k)$
$=i(2-2\lambda )+j(3+4\lambda )+k(1-\lambda )$
$\sin ce\overrightarrow{a_2}and\overrightarrow{b}areperpendicular$
$\overrightarrow{a_2}\cdot \overrightarrow{b}=0$
$[i(2-2\lambda )+j(3+4\lambda )+k(1-\lambda )]\cdot (2i-4j+k)=0$
$\Rightarrow 2(2-2\lambda )+(-4)(3+4\lambda )+(1-\lambda )\left(1\right)=0$
$\Rightarrow 4-4\lambda -12-16\lambda +1-\lambda =0$
$\Rightarrow -7=21\lambda$
$\lambda =\frac{-1}{3}$
${\overrightarrow{a}}_1=\frac{-1}{3}(2i-4j+k)$
${\overrightarrow{a}}_2=i(2-2\left(\frac{-1}{3}\right))+j(3+4\left(\frac{-1}{3}\right))+k(1-\left(\frac{-1}{3}\right))$
$=\frac{1}{3}(8i+5j+4k)$
$Hence$
$\frac{1}{3}(8i+5j+4k)-\frac{1}{3}(2i-4j+k)$
$comparingwithp(8i+5j+4k)-q(2i-4j+k)$
$p=\frac{1}{3},q=\frac{-1}{3}$
$Hence$
$p+q=\frac{1}{3}-\frac{1}{3}=0$