$\begin{aligned} & \text { If } \mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=2…

1. $\left(\frac{11}{5}, \frac{4}{5}, \frac{19}{5}\right)$
2. $\left(\frac{4}{5}, \frac{11}{5}, \frac{19}{5}\right)$
3. $\left(\frac{4}{5}, \frac{19}{5}, \frac{11}{5}\right)$
4. $\left(\frac{19}{5}, \frac{11}{5}, \frac{4}{5}\right)$

Solution

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k}, \mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ $\begin{aligned} \text { Now, } \mathbf{a} \times \mathbf{b} & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 2 & -1 & 3\end{array}\right| \\ & =\hat{\mathbf{i}}(3+1)-\hat{\mathbf{j}}(3-2)+\hat{\mathbf{k}}(-1-2)=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}\end{aligned}$ and $\mathbf{b} \times \mathbf{c}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 3 \\ 1 & -1 & 0\end{array}\right|$ $=\hat{\mathbf{i}}(0+3)-\hat{\mathbf{j}}(0-3)+\hat{\mathbf{k}}(-2+1)=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{c} \times \mathbf{a}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 1 & 1\end{array}\right|$ $ =\hat{\mathbf{i}}(-1-0)-\hat{\mathbf{j}}(1-0)+\hat{\mathbf{k}}(1+1)=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} $ Now, $6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}=\lambda_1(\mathbf{a} \times \mathbf{b})+\lambda_2(\mathbf{b} \times \mathbf{c})$ $ \begin{aligned} \Rightarrow 6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}= & \lambda_1(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \quad+\lambda_3(\mathbf{c} \times \mathbf{a}) \\ & \lambda_2(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda_3(-\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \Rightarrow 6 \hat{\mathbf{i}}+2 & \hat{\mathbf{j}}+3 \hat{\mathbf{k}}=\hat{\mathbf{i}}\left(4 \lambda_1+3 \lambda_2-\lambda_3\right) \\ & +\hat{\mathbf{j}}\left(-\lambda_1+3 \lambda_2-\lambda_3\right)+\hat{\mathbf{k}}\left(-3 \lambda_1-\lambda_2+2 \lambda_3\right) \end{aligned} $ by comparing
Eqs. (i) and (ii), we get $ 5 \lambda_1=4 \Rightarrow \lambda_1=\frac{4}{5} $ put in Eqs. (ii) and (iii), we get $ 3 \lambda_2-\lambda_3=2+\frac{4}{5} $
$ \text { and }-\lambda_2+2 \lambda_3=3+3 \cdot \frac{4}{5} \Rightarrow-\lambda_2+2 \lambda_3=\frac{27}{5} $ multiply by 3 .
Add Eqs. (iv) and (v) $ \begin{aligned} & 5 \lambda_3=\frac{95}{5} \Rightarrow 5 \lambda_3=19 \\ & \lambda_3=\frac{19}{5} \end{aligned} $ put in Eq. (iv) $ \begin{aligned} 3 \lambda_2 & =\frac{14}{5}+\frac{19}{5} \Rightarrow \lambda_2=\frac{33}{15} \\ \lambda_2 & =\frac{11}{5}=\left(\lambda_1, \lambda_2, \lambda_3\right)=\left(\frac{4}{5}, \frac{11}{5}, \frac{19}{5}\right) \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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