Q74 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics
Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ be three vectors. Let $\vec{r}$ be a vector perpendicular to both $\vec{b}, \vec{c}$ and $\vec{r}, \vec{a}=11$. Then the vector among the following that is perpendicular to $\vec{r}$ is
- A$\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
- Correct answerB$\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
- C$\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$
- D$\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$
Solution
Vector perpendicular to $\vec{b}$ and $\vec{c}=\lambda(\vec{b} \times \vec{c})$
$\begin{aligned}<br />& \lambda(\vec{b} \times \vec{c})=\lambda\left|\begin{array}{ccc}<br />\hat{i} & \hat{j} & \hat{k} \\<br />3 & -1 & 5 \\<br />1 & -4 & -2<br />\end{array}\right| \\<br />& =\lambda(22 \hat{i}+11 \hat{j}-11 \hat{k}) \\<br />& \vec{r}=11 \lambda(2 \hat{i}+\hat{j}-\hat{k}) \\<br />& \vec{r} \cdot \vec{a}=11 \\<br />& \Rightarrow 11 \lambda(2 \hat{i}+\hat{j}-\hat{k})(\hat{i}+2 \hat{j}+3 \hat{k})=11 \\<br />& \Rightarrow \lambda(2+2-3)=1 \\<br />& \therefore \lambda=1 \\<br />& \therefore \vec{r}=11(2 \hat{i}+\hat{j}-\hat{k})<br />\end{aligned}$
Vector perpendicular to $\vec{r}$ such that $\vec{r} \cdot \vec{p}=0$
$\Rightarrow(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})=0$
$\therefore$ Vector is $(\hat{i}-\hat{j}+\hat{k})$.
$\begin{aligned}<br />& \lambda(\vec{b} \times \vec{c})=\lambda\left|\begin{array}{ccc}<br />\hat{i} & \hat{j} & \hat{k} \\<br />3 & -1 & 5 \\<br />1 & -4 & -2<br />\end{array}\right| \\<br />& =\lambda(22 \hat{i}+11 \hat{j}-11 \hat{k}) \\<br />& \vec{r}=11 \lambda(2 \hat{i}+\hat{j}-\hat{k}) \\<br />& \vec{r} \cdot \vec{a}=11 \\<br />& \Rightarrow 11 \lambda(2 \hat{i}+\hat{j}-\hat{k})(\hat{i}+2 \hat{j}+3 \hat{k})=11 \\<br />& \Rightarrow \lambda(2+2-3)=1 \\<br />& \therefore \lambda=1 \\<br />& \therefore \vec{r}=11(2 \hat{i}+\hat{j}-\hat{k})<br />\end{aligned}$
Vector perpendicular to $\vec{r}$ such that $\vec{r} \cdot \vec{p}=0$
$\Rightarrow(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})=0$
$\therefore$ Vector is $(\hat{i}-\hat{j}+\hat{k})$.
Hence, the correct answer is option: 'B'.