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Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ be the position vector of a point $\mathrm{A}$. Let $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be two vectors and $\vec{r}$ be a vector passing through the point $A(\vec{a})$ and parallel to the vector $\vec{b}$. If the projection of $\vec{r}$ on $\vec{c}$ is $\frac{9}{\sqrt{6}}$ then $|\vec{r}|=$
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Verified Answer
The correct answer is:
$\sqrt{26}$
Equation of vector passing through $\vec{a}$ and parallel
to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$
$\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})$
$=(2+\lambda) \hat{i}+(2 \lambda-1) \hat{j}+(1-\lambda) \hat{k}$
Projection of $\vec{r}$ on $\vec{c}=\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{9}{\sqrt{6}}$
$\begin{aligned} & \frac{(2+\lambda)+(2 \lambda-1)-2(1-\lambda)}{\sqrt{6}}=\frac{9}{\sqrt{6}} \\ & \Rightarrow 5 \lambda-1=9 \Rightarrow \lambda=2 \\ & \therefore \vec{r}=4 \hat{i}+3 \hat{i}-\hat{k} \\ & |\vec{r}|=\sqrt{4^2+3^2+(-1)^2}=\sqrt{26}\end{aligned}$
to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$
$\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})$
$=(2+\lambda) \hat{i}+(2 \lambda-1) \hat{j}+(1-\lambda) \hat{k}$
Projection of $\vec{r}$ on $\vec{c}=\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{9}{\sqrt{6}}$
$\begin{aligned} & \frac{(2+\lambda)+(2 \lambda-1)-2(1-\lambda)}{\sqrt{6}}=\frac{9}{\sqrt{6}} \\ & \Rightarrow 5 \lambda-1=9 \Rightarrow \lambda=2 \\ & \therefore \vec{r}=4 \hat{i}+3 \hat{i}-\hat{k} \\ & |\vec{r}|=\sqrt{4^2+3^2+(-1)^2}=\sqrt{26}\end{aligned}$
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