# Let the function

Question:

Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \quad \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\left(\lambda^{2}-1\right) \hat{\mathrm{k}}$ be coplanar vectors. Then the non-zero vector $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}$ is :

1. $-14 \hat{i}-5 \hat{j}$

2. $-10 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}$

3. $-10 \hat{i}+5 \hat{j}$

4. $-14 \hat{i}+5 \hat{j}$

Correct Option: , 3

Solution:

$\left[\begin{array}{lll}\overrightarrow{\mathrm{a}} & \overrightarrow{\mathrm{b}} & \overrightarrow{\mathrm{c}}\end{array}\right]=0$

$\Rightarrow\left|\begin{array}{ccc}1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \lambda^{2}-1\end{array}\right|=0$

$\Rightarrow \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0$

$\Rightarrow \lambda^{2}(\lambda-2)-9(\lambda-2)=0$

$\Rightarrow(\lambda-3)(\lambda+3)(\lambda-2)=0$

$\Rightarrow \lambda=2,3,-3$

So, $\lambda=2$ (as $\overrightarrow{\mathrm{a}}$ is parallel to $\overrightarrow{\mathrm{c}}$ for $\lambda=\pm 3$ )

Hence $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 4 \\ 2 & 4 & 3\end{array}\right|$

$=-10 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}$