Question

Let $\overrightarrow{a}=\hat{i}+2\hat{j}-\widehat{k,}\overrightarrow{b}=\hat{i}-\hat{j}$ and $\overrightarrow{c}=\hat{i}-\hat{j}-\hat{k}$be three given vectors. If $\overrightarrow{r}$is a vector such that $\begin{array}{l}\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\end{array}$and $\begin{array}{l}\overrightarrow{r}.\overrightarrow{b}=0\end{array}$then $\begin{array}{l}\overrightarrow{r}.\overrightarrow{a}\end{array}$is equal to _____

Answer 1

Determine the value of $\begin{array}{l}\overrightarrow{r}.\overrightarrow{a}\end{array}$

Step 1: Rearranging the equation $\begin{array}{l}\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\end{array}$

$$\begin{gathered} \Rightarrow \begin{array}{l}\overrightarrow{r}\times \overrightarrow{a}-\overrightarrow{c}\times \overrightarrow{a}=0\end{array} \\ \Rightarrow \left(\begin{array}{l}\overrightarrow{r}-\overrightarrow{c}\end{array}\right)\times \overrightarrow{a}=0....\left(i\right) \end{gathered}$$

If cross product of two vectors are zero, then these two vectors are parallel

So, $\left(\begin{array}{l}\overrightarrow{r}-\overrightarrow{c}\end{array}\right)\&\overrightarrow{a}$are parallel vectors

$$\begin{gathered} \therefore \overrightarrow{r}-\overrightarrow{c}=\lambda \overrightarrow{a} \\ \Rightarrow \overrightarrow{r}=\lambda \overrightarrow{a}+\overrightarrow{c}......\left(ii\right) \\ \Rightarrow \overrightarrow{r}·\overrightarrow{b}=\lambda \overrightarrow{a}·\overrightarrow{b}+\overrightarrow{c}·\overrightarrow{b}[takingdotproductwith\overrightarrow{b]} \end{gathered}$$

Step 2: Finding value of $\lambda$ $$\begin{gathered} \Rightarrow \lambda \left[\left(\hat{i}+2\hat{j}-\hat{k}\right)\left(\hat{i}-\hat{j}\right)\right]+\left[\left(\hat{i}-\hat{j}-\hat{k}\right)\left(\hat{i}-\hat{j}\right)\right]=0\mathbf{}\left[\because \overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}-\hat{j},\overrightarrow{c}=\hat{i}-\hat{j}-\hat{k},\overrightarrow{r}.\overrightarrow{b}=0\&\overrightarrow{r}.\overrightarrow{a}=0aregiven\begin{array}{l}\end{array}\right] \\ \Rightarrow \lambda \left(\hat{i}.\hat{i}+2\hat{j}·\hat{i}-\hat{k}·\hat{i}+\hat{i}·\hat{j}+2\hat{j}·\hat{j}+\hat{k}·\hat{i}\right)+\left(\hat{i}.\hat{i}-\hat{i}.\hat{j}-\hat{j}.\hat{i}+\hat{j}.\hat{j}-\hat{k}.\hat{i}+\hat{k}.\hat{j}\right)=0 \\ \Rightarrow \lambda \left(1-2\right)+2=0\mathbf{}\left[\because \hat{i},\hat{j},\hat{k}\text{are unit vectors},\therefore \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1and\hat{j}.\hat{k}=\hat{k}.\hat{j}=\hat{k}.\hat{i}=\hat{i}.\hat{k}=\hat{i}.\hat{j}=\hat{j}.\hat{i}=0\right] \\ \Rightarrow \lambda =2 \end{gathered}$$

Putting value of $\lambda$ into $\left(ii\right)$

$$\begin{gathered} \Rightarrow \overrightarrow{r}=2\overrightarrow{a}+\overrightarrow{c} \\ \Rightarrow \overrightarrow{r}·\overrightarrow{a}=2\overrightarrow{a}·\overrightarrow{a}+\overrightarrow{c}·\overrightarrow{a}[\text{taking dot produt with}\overrightarrow{a}bothsides] \\ \Rightarrow \overrightarrow{r}·\overrightarrow{a}=2{\left|\overrightarrow{a}\right|}^2+\overrightarrow{c}·\overrightarrow{a} \\ \Rightarrow \overrightarrow{r}·\overrightarrow{a}=2{\left|\overrightarrow{a}\right|}^2+\left(\hat{i}-\hat{j}-\hat{k}\right)\left(\hat{i}+2\hat{j}-\hat{k}\right) \\ \Rightarrow \overrightarrow{r}·\overrightarrow{a}=2{\sqrt{\left(1^2+2^2+1^2\right)}}^2+\left(1-2+1\right)\left[\because \overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}-\hat{j},\&\overrightarrow{c}=\hat{i}-\hat{j}-\hat{k}\right] \\ \Rightarrow \overrightarrow{r}·\overrightarrow{a}=2\times 6+0 \\ \Rightarrow \overrightarrow{r}·\overrightarrow{a}=12 \end{gathered}$$

Hence $\overrightarrow{r}·\overrightarrow{a}=12$is the required answer.