Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2248 Upvotes
Verified Answer
The correct answer is:
$\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
Vector perpendicular to $\vec{b}$ and $\vec{c}=\lambda(\vec{b} \times \vec{c})$
$\begin{aligned}
& \lambda(\vec{b} \times \vec{c})=\lambda\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 5 \\
1 & -4 & -2
\end{array}\right| \\
& =\lambda(22 \hat{i}+11 \hat{j}-11 \hat{k}) \\
& \vec{r}=11 \lambda(2 \hat{i}+\hat{j}-\hat{k}) \\
& \vec{r} \cdot \vec{a}=11 \\
& \Rightarrow 11 \lambda(2 \hat{i}+\hat{j}-\hat{k})(\hat{i}+2 \hat{j}+3 \hat{k})=11 \\
& \Rightarrow \lambda(2+2-3)=1 \\
& \therefore \lambda=1 \\
& \therefore \vec{r}=11(2 \hat{i}+\hat{j}-\hat{k})
\end{aligned}$
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 $(\hat{i}-\hat{j}+\hat{k})$.
$\begin{aligned}
& \lambda(\vec{b} \times \vec{c})=\lambda\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 5 \\
1 & -4 & -2
\end{array}\right| \\
& =\lambda(22 \hat{i}+11 \hat{j}-11 \hat{k}) \\
& \vec{r}=11 \lambda(2 \hat{i}+\hat{j}-\hat{k}) \\
& \vec{r} \cdot \vec{a}=11 \\
& \Rightarrow 11 \lambda(2 \hat{i}+\hat{j}-\hat{k})(\hat{i}+2 \hat{j}+3 \hat{k})=11 \\
& \Rightarrow \lambda(2+2-3)=1 \\
& \therefore \lambda=1 \\
& \therefore \vec{r}=11(2 \hat{i}+\hat{j}-\hat{k})
\end{aligned}$
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 $(\hat{i}-\hat{j}+\hat{k})$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.