Q70 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics

The position vector of a point $P$ is $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{a}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors which determine a plane $\pi$. The equation of a line through $P$ normal to $\mathbf{b}$ and lying on the plane $\pi$ is

  • Correct answer
    A
    $r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-\hat{i}+5 \hat{j}-2 \hat{k})$
  • B
    $r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}+\hat{j}+\hat{k})$
  • C
    $r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-2 \hat{i}-\hat{j}+3 \hat{k})$
  • D
    $r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-3 \hat{i}+4 \hat{j}-5 \hat{k})$

Solution

Given, Position vector of a point $P$ is $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ Equation of plane, $(\mathbf{r}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\mathbf{a} \times \mathbf{b})=0$ $\therefore$ Normal of plane $=\mathbf{a} \times \mathbf{b}$ Equation of line through $\mathbf{P}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ and normal to $\mathbf{b}$ and lying on the plane is $\begin{aligned} & \mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda((\mathbf{a} \times \mathbf{b}) \times \mathbf{b}) \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda((\mathbf{b} . \mathbf{a}) \mathbf{b}-(\mathbf{b} . \mathbf{b}) \mathbf{a}) \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda[(-5)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-6(-\hat{\mathbf{i}}-2 \hat{\mathbf{k}})] \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \end{aligned}$

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

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