Q69 / 160 · AP EAMCET 2016 · Mathematics

The cartesian equation of the plane whose vector equation is $\gamma=(1+\lambda-\mu) \hat{\mathbf{i}}+(2-\lambda) \hat{\mathbf{j}}+(3-2 \lambda+2 \mu) \hat{\mathbf{k}}$ where $\lambda, \mu$ are scalars, is

  • A
    $2 x+y=5$
  • B
    $2 x-y=5$
  • C
    $2 x-z=5$
  • Correct answer
    D
    $2 x+z=5$

Solution

$\begin{aligned} & \text { } \gamma=(1+\lambda-\mu) \hat{i}+(2-\lambda) \hat{j}+(3-2 \lambda+2 \mu) \hat{k} \\ & =\hat{i}+\lambda \hat{i}-\mu \hat{i}+2 \hat{j}-\lambda \hat{j}+3 \hat{k}-2 \lambda \hat{k}+2 \mu \hat{k} \\ & =(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}-2 \hat{k})+\mu(-\hat{i}+2 \hat{k}) \end{aligned}$ Equation of plane is given by $\{r-(\hat{i}+2 \hat{j}+3 \hat{k})\}+[(\hat{i}-\hat{j}-2 \hat{k}) \times(-\hat{i}+2 \hat{k})]=0$ $\begin{array}{ll} \Rightarrow & \{r-(\hat{i}+2 \hat{j}+3 \hat{k})\} \cdot(-2 \hat{i}-\hat{k})=0 \\ \Rightarrow & r(2 \hat{i}+\hat{k})-5=0 \\ \Rightarrow & r(2 \hat{i}+\hat{k})=5 \end{array}$ Now, equation of plane in cartesian form $\begin{aligned} & (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{j}+\hat{k})=5 \\ & \therefore \quad 2 x+z=5 \end{aligned}$

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

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