A unit vector coplanar with $\mathbf{i}+\mathbf{j}+3 \mathbf{k}$ and $\mathbf{i}+3 \mathbf{j}+\mathbf{k}$ and perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$ is

Question

Aicharya · Accepted Answer

Let the unit vector be $ \begin{aligned} & \qquad \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \ & 	ext { and } \mathbf{a}=\mathbf{i}+\mathbf{j}+3 \mathbf{k}, \mathbf{b}=\mathbf{i}+3 \mathbf{j}+\mathbf{k} \ & 	ext { and } \mathbf{c}=\mathbf{i}+\mathbf{j}+\mathbf{k} \end{aligned} $ Given, $[\mathbf{r}, \mathbf{a} \mathbf{b}]=0$, i.e., coplanar. $ \begin{array}{ccc} \Rightarrow & \left|\begin{array}{ccc} x & y & z \ 1 & 1 & 3 \ 1 & 3 & 1 \end{array}ight|=0 \ \Rightarrow & x(1-9)-y(1-3)+z(3-1)=0 \ \Rightarrow & -8 x+2 y+2 x=0 \ \Rightarrow & -4 x+y+z=0 \end{array} $ and $\quad \mathbf{r} \cdot \mathbf{c}=0$, i.e., perpendicular $ \begin{array}{cc} \Rightarrow & (x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) \cdot(\mathbf{i}+\mathbf{j}+\mathbf{k})=0 \ \Rightarrow & x+y+z=0 \end{array} $ On solving Eqs. (i) and (ii), we get $ \begin{aligned} 5 y+5 z & =0 \ y & =-z \end{aligned} $ $\because \mathbf{r}$ is a unit vector. $ \begin{array}{ll} 	herefore & |\mathbf{r}|=1=\sqrt{x^2+y^2+z^2} \ \Rightarrow & x^2+y^2+z^2=1 \ \Rightarrow & x^2+2 y^2=1 	ext { [from Eq. (iii)]. } \end{array} $ Put $y=-z$ in Eq. (i), we get $ -4 x=0 \Rightarrow x=0 $ From Eq. (iv), we get $ 2 y^2=1 \Rightarrow y= \pm \frac{1}{\sqrt{2}} $ Required vector is $ \begin{aligned} \mathbf{r} & =x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \ & =0 \mathbf{i} \mp \frac{1}{\sqrt{2}} \mathbf{j} \pm \frac{1}{\sqrt{2}} \mathbf{k} \ & =\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}} 	ext { or } \frac{-\mathbf{j}+\mathbf{k}}{\sqrt{2}} \end{aligned} $

A unit vector coplanar with $\mathbf{i}+\mathbf{j}+3 \mathbf{k}$ and $\mathbf{i}+3 \mathbf{j}+\mathbf{k}$…

Solution