Question

The dot product of a vector with vectors $\hat{i}+\hat{j}-3\hat{k}$, $\hat{i}+3\hat{j}-2\hat{k}$ ,$2\hat{i}+\hat{j}+4k$ are $0,5$ and $8$ respectively. Find the vector .

Answer 1

Let vertor be $a\hat{i}+b\hat{j}+c\hat{k}$

Dot product with $\hat{i}+\hat{j}-3\hat{k}=a+b-3c=0$ …………$\left(1\right)$

Dot product with $\hat{i}+3\hat{j}-2\hat{k}=a+3b-2c=5$ ……….$\left(2\right)$

Dot product with $2\hat{i}+\hat{j}+4\hat{k}=2a+b+4c=8$ ……………$\left(3\right)$

Equation $\left(2\right)$ -Equation $\left(1\right)$

$\Rightarrow a+3b-2c-a-b+3c=5-0$

$\Rightarrow 2b+c=5$ ………..$\left(4\right)$

Equation $\left(3\right)$-$2\times$ equation $\left(2\right)$

$\Rightarrow 2a+b+4c-2a-6b+4c=8-10$

$-5b+8c=-2$ …………..$\left(5\right)$

Equation $\left(5\right)$ - $4\times$ equation $\left(4\right)$

$\Rightarrow -5b+8c-16b-8c=-2-40$

$\Rightarrow -21b=-42$

$\therefore b=\frac{-42}{-21}=2$

Put $b=2$ in $\left(4\right)$ we get $c=5-2b$

$c=5-2\times 2=5-4$

$\therefore c=1$

Put $b=2$, $c=1$ in $\left(1\right)$

$a+2-3\times 1=0$

$a+2-3=0$

$\therefore a-1=0$

(or) $a=1$

Hence the verctor is $\hat{i}+2\hat{j}+\hat{k}$.