Question

If $a=i-j,b=j-k,c=k-i.$ If $d$ is a unit vector such that $a.d=0=[bcd],$ then $d$ is (are) equal to $?$

• A. $\pm \frac{(i+j-k)}{\sqrt{3}}$
• B. $\pm \frac{(i+j-2k)}{\sqrt{6}}$
• C. $\pm \frac{(i+j+k)}{\sqrt{3}}$
• D. $\pm k$
• E. $i+j$

Answer 1

The correct option is B

$\pm \frac{(i+j-2k)}{\sqrt{6}}$

Explanation for the correct option:

Step 1. Expressing the given data:

Given,

$a=i-j,$

$b=j-k$ and

$c=k-i.$

Now we are given, $[\mathrm{b}\mathrm{c}\mathrm{d}]=0$

So $b,c$ and $d$ are Coplanar vectors

Therefore,

$$\begin{gathered} d=b+\lambda c \\ =(\mathrm{j}-\mathrm{k})+\lambda (-\mathrm{i}+\mathrm{k}) \end{gathered}$$

$\Rightarrow$$d=j-k-\lambda i+\lambda k$

$\Rightarrow$$d=-\lambda i+j+\lambda k-k$

$\Rightarrow$$d=-\lambda i+j+\left(\lambda -1\right)k$

$\Rightarrow$$d=(-\mathrm{\lambda })i+j+(\mathrm{\lambda }-1)k\to \left(i\right)$

Now,

$a.d=0$

$\Rightarrow$ $\left(i-j\right).\left[\left(-\lambda \right)i+j+\left(\lambda -1\right)k\right]=0$

$\Rightarrow$$-\lambda i.i+\lambda i.j+i.j-j.j+\left(\lambda k-k\right).i-\left(\lambda k-k\right).j=0$

$\Rightarrow$ $-\lambda -1+0=0$ $\left[\mathbf{\because }\mathit{i}\mathbf{·}\mathit{j}\mathbf{}\mathbf{=}\mathbf{}\mathit{j}\mathbf{·}\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathit{k}\mathbf{·}\mathit{i}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\right]$

$\Rightarrow$ $-\lambda -1=0$

$\Rightarrow$ $-\lambda =1$

Multiply negative on both side

$\Rightarrow$ $\lambda =-1$

Step 2. Put $\lambda =-1$ in equation $\left(i\right)$

$d=(-\mathrm{\lambda })i+j+(\mathrm{\lambda }-1)k$

$d=\left(-\left(-1\right)\right)i+j+\left(-1-1\right)k$

$\Rightarrow$ $d=1i+j+\left(-2\right)k$

$\Rightarrow$ $d=i+j-2k$

Step 3. Finding the value of $\hat{d}$.

Now ,$\hat{d}=\pm \frac{d}{\left|d\right|}$ $\left[\mathbf{\because }\left|d\right|\mathbf{}\mathbf{=}\mathbf{}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\right]$

$\Rightarrow$ $\hat{d}=\frac{\pm \left(i+j-2k\right)}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-2\right)}^2}}$

$\Rightarrow$ $\hat{d}=\frac{\pm \left(i+j-2k\right)}{\sqrt{1+1+4}}$

$\Rightarrow$ $\hat{\mathbf{d}}\mathbf{=}\frac{\mathbf{\pm }\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{j}\mathbf{-}\mathbf{2}\mathbf{k}\mathbf{)}}{\sqrt{\mathbf{6}}}$

Hence , option (B) is correct.