Question

Find the values of x and y if the vectors $\overrightarrow{a}=3\hat{i}+x\hat{j}-\hat{k}\mathrm{and}\overrightarrow{b}=2\hat{i}+\hat{j}+y\hat{k}$ are mutually perpendicular vectors of equal magnitude.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have} \\ \overrightarrow{a}=3\hat{i}+x\hat{j}-\hat{k}\mathrm{and}\overrightarrow{b}=2\hat{i}+\hat{j}+y\hat{k} \\ \text{It is given that the vectors are perpendicular.} \\ \Rightarrow \overrightarrow{a}.\overrightarrow{b}=0 \\ \Rightarrow 6+x-y=0 \\ \Rightarrow x-y=-6...\left(1\right) \\ \text{Also, it is given that} \\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right| \\ \Rightarrow \sqrt{9+x^2+1}=\sqrt{4+1+y^2} \\ \Rightarrow \sqrt{10+x^2}=\sqrt{5+y^2} \\ \Rightarrow 10+x^2=5+y^2 \\ \Rightarrow x^2-y^2=-5 \\ \Rightarrow \left(x+y\right)\left(x-y\right)=-5 \\ \Rightarrow -6\left(x+y\right)=-5\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow x+y=\frac{5}{6}...\left(2\right) \\ \text{(1)+(2) gives} \\ 2x=\frac{-31}{6} \\ \Rightarrow x=\frac{-31}{12} \\ \text{From (1),} \\ \frac{-31}{12}-y=-6 \\ \Rightarrow y=\frac{-31}{12}+6=\frac{41}{12} \\ \therefore x=\frac{-31}{12}\mathrm{and}y=\frac{41}{12} \end{gathered}$$