Question

If the dot product of a vector with vectors $\hat{i}-\hat{j},\hat{i}+\hat{k}$ and $\hat{i}+\hat{j}-2\hat{k}$ are respectively $1,-6$ and $3$, then the vector is

• A. $\frac{-5}{2}\hat{i}-\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k}$
• B. $3\hat{j}-4\hat{k}$
• C. $2\hat{i}-4\hat{k}$
• D. $-2\hat{i}-3\hat{j}-4\hat{k}$

Answer 1

The correct option is A $\frac{-5}{2}\hat{i}-\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k}$

Let the unknown vector be $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$

and $\overrightarrow{b}=\hat{i}-\hat{j}$

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

$\overrightarrow{d}=\hat{i}+\hat{j}-2\hat{k}$

According to the question,

$\overrightarrow{a}.\overrightarrow{b}=(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}-\hat{j})=1$

$\Rightarrow x-y=1$ ........$\left(1\right)$

$\overrightarrow{a}.\overrightarrow{c}=(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+\hat{j})=-6$

$\Rightarrow x+y=-6$ ........$\left(2\right)$

$\overrightarrow{a}.\overrightarrow{d}=(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+\hat{j}-2\hat{k})=3$

$\Rightarrow x+y-2z=3$ ........$\left(3\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$ we get

$x-y+x+y=1-6$

$\Rightarrow 2x=-5$

$\therefore x=\frac{-5}{2}$

Put $x=\frac{-5}{2}$ in $x+y=-6$

$\Rightarrow y=-6+\frac{5}{2}=\frac{-12+5}{2}=\frac{-7}{2}$

Put the values of $x=\frac{-5}{2}$ and $y=\frac{-7}{2}$ in $x+y-2z=3$ we get

$\frac{-5}{2}-\frac{7}{2}-2z=3$

$\Rightarrow -6-2z=3$

$\Rightarrow -2z=3+6=9$

$\therefore z=\frac{-9}{2}$

$\therefore$ the vector is $\frac{-5}{2}\hat{i}-\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k}$