Q80 / 160 · AP EAMCET 2004 · Mathematics

If $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}, \sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are coplanar, then $\lambda$ is equal to

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    4

Solution

Let $\mathbf{a}=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $$ \mathbf{c}=\sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}} $$ Since, these vectors are coplanar $$ \begin{aligned} & \therefore \quad \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} 3 & 3 & \sqrt{3} \\ 1 & 0 & 1 \\ \sqrt{3} & \sqrt{3} & \lambda \end{array}\right|=0 \\ & \Rightarrow \quad 3(0-\sqrt{3})-3(\lambda-\sqrt{3})+\sqrt{3}(\sqrt{3}-0)=0 \\ & \Rightarrow \quad-3 \sqrt{3}-3 \lambda+3 \sqrt{3}+3=0 \\ & \Rightarrow \quad \lambda=1 \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.

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