Let vec{a}=2 hat{i}-hat{j}+hat{k} be the posi | vedclass

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(A) The equation of the line passing through point $A$ with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a}

Vedclass · Accepted Answer

$\sqrt{26}$

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(A) The equation of the line passing through point $A$ with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.Substituting the given vectors: $\vec{r} = (2 \hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + 2 \hat{j} - \hat{k}) = (2+\lambda) \hat{i} + (2\lambda-1) \hat{j} + (1-\lambda) \hat{k}$.The projection of $\vec{r}$ on $\vec{c}$ is given by $\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|} = \frac{9}{\sqrt{6}}$.First,calculate $|\vec{c}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$.Now,calculate the dot product $\vec{r} \cdot \vec{c} = (2+\lambda)(1) + (2\lambda-1)(1) + (1-\lambda)(-2) = 2 + \lambda + 2\lambda - 1 - 2 + 2\lambda = 5\lambda - 1$.Equating the projection: $\frac{5\lambda - 1}{\sqrt{6}} = \frac{9}{\sqrt{6}} \Rightarrow 5\lambda - 1 = 9 \Rightarrow 5\lambda = 10 \Rightarrow \lambda = 2$.Substituting $\lambda = 2$ into the expression for $\vec{r}$: $\vec{r} = (2+2) \hat{i} + (2(2)-1) \hat{j} + (1-2) \hat{k} = 4 \hat{i} + 3 \hat{j} - \hat{k}$.Finally,the magnitude $|\vec{r}| = \sqrt{4^2 + 3^2 + (-1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$.

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