Question

Let a=i-j,b=j-k,c=k-i and d is a unit vector such that a.d=0=[bcd].Find d.

Answer 1

Therefore the value  $\overrightarrow{d}=\sqrt{\frac{1}{6} } i+\sqrt{\frac{1}{6} } j-2\sqrt{\frac{1}{6} } k$  and $\overrightarrow{d}=-\sqrt{\frac{1}{6} } i-\sqrt{\frac{1}{6} } j+2\sqrt{\frac{1}{6} } k$

Step-by-step explanation:

Given;

$\overrightarrow{a}=i+j$ ,  $\overrightarrow{b}=j-k$ and $\overrightarrow{c}=k-i$

Let,

$\overrightarrow{d}=pi+qj+rk$

From question,

   $\overrightarrow{a}\times \overrightarrow{d}=0$

⇒$(i-j)\times (pi+qj+rk)$$=0$

⇒$p-q=0$

⇒$p=q$

Then,

$\overrightarrow{d}=pi+pj+rk$

And also,

  $\overrightarrow{b}.\overrightarrow{c}.\overrightarrow{d}=0$

⇒$\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{d})=0$

$\overrightarrow{c}\times \overrightarrow{d}=(k-i)\times (pi+pj+rk)$$= i(-p)-j(-r-p)+k(-p)$$= -pi+(r+p)j-pk$

⇒$\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{d})=(j-k).(-pi+(r+p)j-pk)$$=0$

⇒$r+p+p=0$

⇒$r=-2p$

We know,

$\left |\overrightarrow{d}\right |=\sqrt{p^{2} +q^{2} +r^{2}} =1$

⇒$p^{2} +q^{2} +r^{2} =1$

⇒$p^{2} +p^{2} +4p^{2} =1$

⇒$p^{2} =\frac{1}{6}$

Then,

$p=\sqrt{ \frac{1}{6}}$  or $p=-\sqrt{ \frac{1}{6}}$

If $p=\sqrt{ \frac{1}{6}}$  then $q=\sqrt{ \frac{1}{6}}$ , $r=-2\sqrt{ \frac{1}{6}}$

and $p=-\sqrt{ \frac{1}{6}}$ then $q=-\sqrt{ \frac{1}{6}}$ , $r=2\sqrt{ \frac{1}{6}}$

Then,

The value  $\overrightarrow{d}=\sqrt{\frac{1}{6} } i+\sqrt{\frac{1}{6} } j-2\sqrt{\frac{1}{6} } k$  and $\overrightarrow{d}=-\sqrt{\frac{1}{6} } i-\sqrt{\frac{1}{6} } j+2\sqrt{\frac{1}{6} } k$