Question

Let $\stackrel{\rightharpoonup }{a}=\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}-3\stackrel{\rightharpoonup }{k}$ and $\stackrel{\rightharpoonup }{b}=2\stackrel{\rightharpoonup }{i}-3\stackrel{\rightharpoonup }{j}+5\stackrel{\rightharpoonup }{k}$. If $\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{a}=\stackrel{\rightharpoonup }{b}\times \overrightarrow{r}$, $\stackrel{\rightharpoonup }{r}.\left(a\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}+\stackrel{\rightharpoonup }{k}\right)=3$ and $\stackrel{\rightharpoonup }{r}.\left(2\stackrel{\rightharpoonup }{i}+5\stackrel{\rightharpoonup }{j}-a\stackrel{\rightharpoonup }{k}\right)=-1$,$ɑ\in R$, then the value of $a+{\left|\stackrel{\rightharpoonup }{r}\right|}^2=$

• A. $11$
• B. $15$
• C. $9$
• D. $13$

Answer 1

The correct option is B

$15$

Step 1: Use cross product of parallel vectors is zero

Let $\stackrel{\rightharpoonup }{r}=x\stackrel{\rightharpoonup }{i}+y\stackrel{\rightharpoonup }{j}+z\stackrel{\rightharpoonup }{k}$

Given,$\stackrel{\rightharpoonup }{a}=\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}-3\stackrel{\rightharpoonup }{k}$ , $\stackrel{\rightharpoonup }{b}=2\stackrel{\rightharpoonup }{i}-3\stackrel{\rightharpoonup }{j}+5\stackrel{\rightharpoonup }{k}$

$$\begin{gathered} \Rightarrow \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{a}=\stackrel{\rightharpoonup }{b}\times \overrightarrow{r} \\ \Rightarrow \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{a}=-\overrightarrow{r}\times \stackrel{\rightharpoonup }{b} \\ \Rightarrow \left(\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{a}\right)+\left(\overrightarrow{r}\times \stackrel{\rightharpoonup }{b}\right)=0 \\ \Rightarrow \stackrel{\rightharpoonup }{r}\times \left(\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right)=0 \end{gathered}$$

$\therefore \stackrel{\rightharpoonup }{r}\parallel \left(\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right)$since$\stackrel{\rightharpoonup }{r}\times \left(\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right)=0$ and two vectors are in parallel if and only if the cross vector of that two vectors is zero

$\begin{array}{rcl}Let\stackrel{\rightharpoonup }{r}& =& \lambda \left(\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right)\\ & =& \lambda \left[\left(\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}-3\stackrel{\rightharpoonup }{k}\right)+2\stackrel{\rightharpoonup }{i}-3\stackrel{\rightharpoonup }{j}+5\stackrel{\rightharpoonup }{k}\right]\\ & =& \lambda \left[3\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j}+2\stackrel{\rightharpoonup }{k}\right]\end{array}$

Step 2: Solve given equations to find $\lambda anda$

$$\begin{gathered} \because \stackrel{\rightharpoonup }{r}.\left(a\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}+\stackrel{\rightharpoonup }{k}\right)=3 \\ \Rightarrow \lambda \left(3\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j}+2\stackrel{\rightharpoonup }{k}\right).\left(a\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}+\stackrel{\rightharpoonup }{k}\right)=3 \\ \Rightarrow \lambda \left(3a-2+2\right)=3 \\ \Rightarrow 3a\lambda =3 \\ \Rightarrow \lambda a=1\to \left(i\right) \end{gathered}$$

$$\begin{gathered} \because \stackrel{\rightharpoonup }{r}.\left(2\stackrel{\rightharpoonup }{i}+5\stackrel{\rightharpoonup }{j}-a\stackrel{\rightharpoonup }{k}\right)=-1 \\ \Rightarrow \lambda \left(3\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j}+2\stackrel{\rightharpoonup }{k}\right).\left(2\stackrel{\rightharpoonup }{i}+5\stackrel{\rightharpoonup }{j}-a\stackrel{\rightharpoonup }{k}\right)=-1 \\ \Rightarrow \lambda \left(6-5-2a\right)=-1 \\ \Rightarrow \left(1-2a\right)\lambda =-1\to \left(ii\right) \end{gathered}$$

Solving the equation $\left(i\right)\text{and}\left(ii\right)$

we get

$$\begin{gathered} \left(1-2\frac{1}{\lambda }\right)\lambda =-1\left[\because a=\frac{1}{\lambda }byequ\left(i\right)\right] \\ \lambda -2=-1 \\ \lambda =1\Rightarrow a=1 \end{gathered}$$

Step 3: Determine the value of $a+{\left|\stackrel{\rightharpoonup }{r}\right|}^2$

We get $\stackrel{\rightharpoonup }{r}=\left[3\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j}+2\stackrel{\rightharpoonup }{k}\right]$

thus, $a+{\left|\stackrel{\rightharpoonup }{r}\right|}^2$ is given as:

$a=1$ and $\left|\stackrel{\rightharpoonup }{r}\right|=\sqrt{3^2+{\left(-1\right)}^2+{\left(2\right)}^2}\mathrm{or}\sqrt{14}$

$$\begin{gathered} \therefore a+{\left|\stackrel{\rightharpoonup }{r}\right|}^2=1+14 \\ =15 \end{gathered}$$

Hence, option (B) is the correct answer