Q71 / 180 · KCET 2018 · Mathematics
If \( \vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k} ; \vec{b}=\mu \hat{i}+\hat{j}-\hat{k} \) are orthogonal and \( |\vec{a}|=|\vec{b}| \) then
\( (\lambda, \mu)= \)
\( (\lambda, \mu)= \)
- Correct answerA\( \left(\frac{1}{4}, \frac{7}{4}\right) \)
- B\( \left(\frac{7}{4}, \frac{1}{4}\right) \)
- C\( \left(\frac{1}{4}, \frac{9}{4}\right) \)
- D\( \left(\frac{-1}{4}, \frac{9}{4}\right) \)
Solution
Given vectors,
\[ \begin{array}{l} \vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k} \rightarrow(1) \\ \vec{b}=\mu \hat{i}+\hat{j}-\hat{k} \rightarrow(2) \end{array} \]
For vectors to be orthogonal, we have
\[ \begin{array}{l} \vec{a} \cdot \vec{b}=0 \\ \Rightarrow(\hat{i}+\lambda \hat{j}+2 \hat{k}) \cdot(\mu \hat{i}+\hat{j}-\hat{k})=0 \\ \Rightarrow \mu+\lambda-2=0 \\ \Rightarrow \lambda+\mu=2 \rightarrow(3) \\ \text { and }|\vec{a}|=|\vec{b}| \\ \Rightarrow \sqrt{1+\lambda^{2}+4}=\sqrt{\mu^{2}+1+1} \\ \Rightarrow 1+\lambda^{2}+4=\mu^{2}+1+1 \rightarrow(4) \end{array} \]
From Eqs. (3) and (4), we get
\[ \begin{array}{l} \lambda^{2}+3=\mu^{2}=(2 \lambda)^{2} \\ \Rightarrow \lambda^{2}+3=4+\lambda^{2}-4 \lambda \\ \text { So, } \lambda=\frac{1}{4} \Rightarrow \mu=2-\frac{1}{4}=\frac{7}{4} \end{array} \]
\[ \text { Hence }(\lambda, \mu)=\left(\frac{1}{4}, \frac{7}{4}\right) \]
\[ \begin{array}{l} \vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k} \rightarrow(1) \\ \vec{b}=\mu \hat{i}+\hat{j}-\hat{k} \rightarrow(2) \end{array} \]
For vectors to be orthogonal, we have
\[ \begin{array}{l} \vec{a} \cdot \vec{b}=0 \\ \Rightarrow(\hat{i}+\lambda \hat{j}+2 \hat{k}) \cdot(\mu \hat{i}+\hat{j}-\hat{k})=0 \\ \Rightarrow \mu+\lambda-2=0 \\ \Rightarrow \lambda+\mu=2 \rightarrow(3) \\ \text { and }|\vec{a}|=|\vec{b}| \\ \Rightarrow \sqrt{1+\lambda^{2}+4}=\sqrt{\mu^{2}+1+1} \\ \Rightarrow 1+\lambda^{2}+4=\mu^{2}+1+1 \rightarrow(4) \end{array} \]
From Eqs. (3) and (4), we get
\[ \begin{array}{l} \lambda^{2}+3=\mu^{2}=(2 \lambda)^{2} \\ \Rightarrow \lambda^{2}+3=4+\lambda^{2}-4 \lambda \\ \text { So, } \lambda=\frac{1}{4} \Rightarrow \mu=2-\frac{1}{4}=\frac{7}{4} \end{array} \]
\[ \text { Hence }(\lambda, \mu)=\left(\frac{1}{4}, \frac{7}{4}\right) \]
Hence, the correct answer is option: 'A'.