Question

A unit vector which is coplanar to vector $i+2j+k$ and $i+2j+k$ and perpendicular to $i+j+k$, is

• A. $\frac{i-j}{\sqrt{2}}$
• B. $\pm \left(\frac{j-k}{\sqrt{2}}\right)$
• C. $\frac{k-i}{\sqrt{2}}$
• D. $\frac{i+j+k}{\sqrt{3}}$

Answer 1

The correct option is B $\pm \left(\frac{j-k}{\sqrt{2}}\right)$

Let the vector be given as $aj+bj+ck$. For this vector to be coplanar with $i+j+2k$ and $i+2j+k$ we will have $ai+bj+ck=p(i+j+2k)+r(i+2j+k)$ this gives
$a=p+r....\left(i\right)$
$b=p+2r....\left(ii\right)$
$c=2p+r.....\left(iii\right)$
For the vector $ai+bj+ck$ to be perpendicular to $i+j+k$ we will have
$(ai+bj+ck).(i+j+k)=0$
$\Rightarrow$ $a+b+c=\mathrm{0.....}\left(iv\right)$
On adding Eqs.(i), (ii), (iii) we get
$4p+4r=a+b+c$
$\Rightarrow$ $4(p+r)=0$
$\Rightarrow$ $p=-r$
Now with the help of Eqs (i), (ii) and (iii) we get
$a=0,b=r,c=p=-r$
Hence the required vector is $r(j-k)$
For unit vector, $r^2+r^2=1$
$\Rightarrow r=\pm \frac{1}{\sqrt{2}}$
Hence the required unit vector is $\pm \frac{1}{\sqrt{2}}(j-k)$