If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$, $3 \hat{i}+2 \hat{j}-5 \hat{k},-3…

If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$, $3 \hat{i}+2 \hat{j}-5 \hat{k},-3 \hat{i}+8 \hat{j}-5 \hat{k}$ and $-3 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar, then $\lambda=$
  1. 1
  2. 2
  3. -2
  4. -3

Solution

Let $\overrightarrow{\mathrm{OA}}=-\hat{i}+4 \hat{j}-4 \hat{k} ; \overrightarrow{\mathrm{OB}}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ $\begin{aligned} & \overrightarrow{\mathrm{OC}}=-3 \hat{i}+8 \hat{j}-5 \hat{k} \text { and } \overrightarrow{\mathrm{OD}}=-3 \hat{i}+2 \hat{j}+\lambda \hat{k} \\ & \overrightarrow{\mathrm{AB}}=4 \hat{i}-2 \hat{j}-\hat{k}, \overrightarrow{\mathrm{AC}}=-2 \hat{i}+4 \hat{j}-\hat{k} \\ & \overrightarrow{\mathrm{AD}}=-2 \hat{i}-2 \hat{j}+(\lambda+4) \hat{k} \end{aligned}$
For coplanar $[\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC}} \overrightarrow{\mathrm{AD}}]=0$ $\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 4 & -2 & -1 \\ -2 & 4 & -1 \\ -2 & -2 & \lambda+4 \end{array}\right|=0 \\ & \Rightarrow 4(4 \lambda+16-2)+2(-2 \lambda-8-2)-1(4+8)=0 \\ & \Rightarrow 12 \lambda+24=0 \Rightarrow \lambda=-2 \end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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