Question

If $\overline{a}=4\overline{i}+5\overline{j}-\overline{k},\overline{b}=\overline{i}-4\overline{j}+5\overline{k}$ and $\overline{c}=3\overline{i}+\overline{j}-\overline{k}$ , then find vector$\overline{d}$ which is perpendicular to both $\overline{a},\overline{b}$ and $\overline{d}.\overline{c}=21$

Answer 1

$\overline{a}=4\hat{i}+5\hat{j}-\hat{k},\overline{b}=i-4\hat{j}+5\hat{k}$ and $c=3\hat{i}+\hat{j}-\hat{k}$

Let $\overline{a}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overline{d}$ is perpendicular to a and b

$\therefore \overline{d}.\overline{d}=0$

$(4\hat{i}+5\hat{j}-\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=0$

$4x+5y-z=0$ ...(1)

$\overline{d}$ is perpendicular to b

$\therefore \overline{d}.\overline{b}=0$

$(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}-4\hat{j}+5\hat{k})=0$

$x-4y+5z=0$ ...(2)

$\overline{d}.\overline{c}=21$ (Given)

$(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+\hat{j}-\hat{k})=21$

$3x+y-z=21$ ...(3)

solving (1), (2)

$\frac{x}{25-(-1)(-4)}=\frac{-y}{20-(-1)\left(1\right)}=\frac{z}{16-\left(5\right)\left(1\right)}$

$\frac{x}{21}=\frac{-y}{21}=\frac{z}{11}$

$\therefore x=\frac{z\times 21}{11}$ and $y=\frac{-21z}{11}$ putting in (3)

$\frac{3z\times 21}{11}-\frac{21z}{11}-z=21$

$z=\frac{231}{31}$

$\therefore x=\frac{441}{31}$ and $y=\frac{-441}{31}$

$\overline{d}=\frac{441}{31}\hat{i}-\frac{441}{31}\hat{j}+\frac{231}{31}\hat{k}$