Question

For what value of λ are the vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$perpendicular to each other if
(i) $\overrightarrow{a}=\lambda \hat{i}+2\hat{j}+\hat{k}\mathrm{and}\overrightarrow{b}=4\hat{i}-9\hat{j}+2\hat{k}$

(ii) $\overrightarrow{a}=\lambda \hat{i}+2\hat{j}+\hat{k}\mathrm{and}\overrightarrow{b}=5\hat{i}-9\hat{j}+2\hat{k}$

(iii) $\overrightarrow{a}=2\hat{i}+3\hat{j}+4\hat{k}\mathrm{and}\overrightarrow{b}=3\hat{i}+2\hat{j}-\lambda \hat{k}$

(iv) $\overrightarrow{a}=\lambda \hat{i}+3\hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b}=\hat{i}-\hat{j}+3\hat{k}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\text{If the vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{are perpendicular to each other, then} \\ \overrightarrow{a}.\overrightarrow{b}=0 \\ \Rightarrow \left(\lambda \stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+\stackrel{\wedge }{k}\right).\left(4\stackrel{\wedge }{i}-9\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}\right)=0 \\ \Rightarrow 4\lambda -18+2=0 \\ \Rightarrow 4\lambda -16=0 \\ \Rightarrow 4\lambda =16 \\ \Rightarrow \lambda =4 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\text{If the vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{are perpendicular to each other, then} \\ \overrightarrow{a}.\overrightarrow{b}=0 \\ \Rightarrow \left(\lambda \stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+\stackrel{\wedge }{k}\right).\left(5\stackrel{\wedge }{i}-9\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}\right)=0 \\ \Rightarrow 5\lambda -18+2=0 \\ \Rightarrow 5\lambda -16=0 \\ \Rightarrow 5\lambda =16 \\ \Rightarrow \lambda =\frac{16}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\text{If the vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{are perpendicular to each other, then} \\ \overrightarrow{a}.\overrightarrow{b}=0 \\ \Rightarrow \left(2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}+4\stackrel{\wedge }{k}\right).\left(3\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}-\lambda \stackrel{\wedge }{k}\right)=0 \\ \Rightarrow 6+6-4\lambda =0 \\ \Rightarrow 12-4\lambda =0 \\ \Rightarrow 4\lambda =12 \\ \Rightarrow \lambda =3 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\text{If the vectors}\overrightarrow{a}\text{and}\overrightarrow{b}\text{are perpendicular to each other, then} \\ \overrightarrow{a}.\overrightarrow{b}=0 \\ \Rightarrow \left(\lambda \stackrel{\wedge }{i}+3\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}\right).\left(\stackrel{\wedge }{i}-1\stackrel{\wedge }{j}+3\stackrel{\wedge }{k}\right)=0 \\ \Rightarrow \lambda -3+6=0 \\ \Rightarrow \lambda +3=0 \\ \Rightarrow \lambda =-3 \end{gathered}$$