Question

Dot products of a vector with vectors $3\hat{i}-5\hat{k},2\hat{i}+7\hat{j}$ and $\hat{i}+\hat{j}+\hat{k}$ are respectively - 1, 6 and 5. Find the vector.

Answer 1

Let $\overline{a}=3\hat{i}-5\hat{k},\overline{b}=2\hat{i}+7\hat{j},\overline{c}=\hat{i}+\overline{j}+\hat{k}$

Let $\overline{r}=x\hat{i}+y\overline{j}+z\hat{k}$ be the required vector.

Then, $\overline{r}.\overline{a}=-1,\overline{r}.\overline{a}=6,\overline{r}.\overline{c}=5$

$\therefore (x\hat{i}+y\overline{j}+z\hat{k}).(3\hat{i}-5\hat{k})=-1(x\hat{i}+y\overline{j}+z\hat{k}).(2\hat{i}+7\hat{j})=6$ and

$(x\hat{i}+y\overline{j}+z\hat{k}).(\hat{i}+\hat{j}+\hat{k})=5$

$\therefore 3x-5y=-1$...$\left(1\right)$

$\therefore 2x+7y=6$...$\left(1\right)$

$\therefore x+y+z=5$...$\left(1\right)$

From $\left(3\right),z=5-x-y$

Substituting his value of $z$ in $\left(1\right),$ we get

$\therefore 3x-5(5-x-7)=-1$

$\therefore 8x+5y=24$...$\left(4\right)$

Multiplying $\left(2\right)$ by $4$ and subtracting from $\left(4\right)$, we get

$8x+5y-4(2x+7y)=24-6\times 4\therefore -23y=0\therefore y=0$

Substituting $y=0$ in $\left(2\right),$ we get

$\therefore 2x=6\therefore x=3$

Substituting $x=3$ in $\left(1\right),$ we get

$\therefore 3\left(3\right)-5z=-1\therefore -5z=-10\therefore z=2\therefore \overline{r}=3\hat{i}+0.\hat{j}+2\hat{k}=3\hat{i}+2\hat{k}$

Hence, the required vector is $3\hat{i}+2\hat{k}.$