The vector which is parallel to the resultant | vedclass

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Question

(D) Let $\vec{r}$ be the resultant vector of $\vec{a}$ and $\vec{b}$.
$\vec{r} = \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\h

Vedclass · Accepted Answer

$\frac{15}{\sqrt{14}}\hat{i} + \frac{5}{\sqrt{14}}\hat{j} - \frac{10}{\sqrt{14}}\hat{k}$

Vedclass · Answer

(D) Let $\vec{r}$ be the resultant vector of $\vec{a}$ and $\vec{b}$.
$\vec{r} = \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} - \hat{k}) = 3\hat{i} + \hat{j} - 2\hat{k}$.
The magnitude of $\vec{r}$ is $|\vec{r}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
The unit vector in the direction of $\vec{r}$ is $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{3\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{14}}$.
$A$ vector of magnitude $5$ units parallel to $\vec{r}$ is given by $5\hat{r} = \frac{5}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k}) = \frac{15}{\sqrt{14}}\hat{i} + \frac{5}{\sqrt{14}}\hat{j} - \frac{10}{\sqrt{14}}\hat{k}$.

The vector which is parallel to the resultant vector of $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} - \hat{k}$ and having a magnitude of $5$ units is . . . . . . .

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