Vectors 3 Question 2
2. Let $\quad \mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \quad \mathbf{b}=\hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \quad$ and $\mathbf{c}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\left(\lambda^{2}-1\right) \hat{\mathbf{k}}$ be coplanar vectors. Then, the non-zero vector $\mathbf{a} \times \mathbf{c}$ is
(2019 Main, 11 Jan I)
(a) $-10 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}$
(b) $-10 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}$
(c) $-14 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}$
(d) $-14 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}$
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Answer:
Correct Answer: 2. (a)
Solution:
- We know that, if $a, b, c$ are coplanar vectors, then $\left[\begin{array}{lll}a & b & c\end{array}\right]=0$
$$ \begin{aligned} & \therefore \quad\left|\begin{array}{ccc} 1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \lambda^{2}-1 \end{array}\right|=0 \\ & \Rightarrow 1{\lambda\left(\lambda^{2}-1\right)-16}-2\left(\left(\lambda^{2}-1\right)-8\right)+4(4-2 \lambda)=0 \\ & \Rightarrow \lambda^{3}-\lambda-16-2 \lambda^{2}+18+16-8 \lambda=0 \\ & \Rightarrow \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0 \\ & \Rightarrow \lambda^{2}(\lambda-2)-9(\lambda-2)=0 \end{aligned} $$
$$ \begin{aligned} & \Rightarrow(\lambda-2)\left(\lambda^{2}-9\right)=0 \\ & \Rightarrow(\lambda-2)(\lambda+3)(\lambda-3)=0 \\ & \therefore \lambda=2,3 \text { or }-3 \\ & \text { If } \lambda=2 \text {, then } \\ & \begin{aligned} a \times c=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array}\right|=\hat{i}(6-16)-\hat{j}(3-8)+\hat{k}(4-4) \\ =-10 \hat{i}+5 \hat{j} \end{aligned} \\ & \text { If } \lambda= \pm 3 \text {, then a } \times c=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 8 \end{array}\right|=0 \end{aligned} $$
$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k}\end{array}\right|$
(because last two rows are proportional).
