Question

What is $\overrightarrow{c}$ equal to?

• A. $\frac{3(\hat{i}+\hat{j})}{2}$
• B. $\frac{2(\hat{i}+\hat{j})}{3}$
• C. $\frac{(\hat{i}+\hat{j})}{2}$
• D. $\frac{(\hat{i}+\hat{j})}{3}$

Answer 1

The correct option is A $\frac{3(\hat{i}+\hat{j})}{2}$

THe question has mistake

Here $\overrightarrow{d}$ is perpendicular to $\overrightarrow{a}$

Given

$\overrightarrow{a}=\hat{i}+\hat{j}$

$\Rightarrow \overrightarrow{b}=3\hat{i}+4\hat{k}$

$\overrightarrow{c}$ is parallel to $\overrightarrow{a}$

$\Rightarrow \overrightarrow{c}=\lambda \overrightarrow{a}$

$\Rightarrow \overrightarrow{c}=\lambda (\hat{i}+\hat{j})$

$\Rightarrow \overrightarrow{c}=\lambda \hat{i}+\lambda \hat{j}$

let $\overrightarrow{d}=a\hat{i}+b\hat{j}+c\hat{k}$

$\Rightarrow \overrightarrow{d}$ is perpendicular to $\overrightarrow{a}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{d}=0$

$\Rightarrow (\hat{i}+\hat{j})\cdot (a\hat{i}+b\hat{j}+c\hat{k})=0$

$\Rightarrow a+b=0$

$\Rightarrow a=-b$

putting in $\overrightarrow{d}$

$\Rightarrow \overrightarrow{d}=a\hat{i}-a\hat{j}+c\hat{k}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{c}+\overrightarrow{d}$

$\Rightarrow 3\hat{i}+4\hat{k}=\lambda \hat{i}+\lambda \hat{j}+a\hat{i}-a\hat{j}+c\hat{k}$

$\Rightarrow 3\hat{i}+4\hat{k}=(\lambda +a)\hat{i}+(\lambda -a)\hat{j}+c\hat{k}$

on comparing both sides

$\Rightarrow \lambda +a=3$-----(1)

$\Rightarrow \lambda -a=0,c=4$

$\Rightarrow \lambda =a$

putting value of a in eq (1)

$\Rightarrow \lambda +\lambda =3$

$\Rightarrow 2\lambda =3$

$\Rightarrow \lambda =\frac{3}{2}$

$\Rightarrow \overrightarrow{c}=\lambda (\hat{i}+\hat{j})$

$\Rightarrow \overrightarrow{c}=\frac{3}{2}(\hat{i}+\hat{j})$

$\Rightarrow \overrightarrow{c}=\frac{3(\hat{i}+\hat{j})}{2}$

Hence option A is correct