Question

If the vector $\overrightarrow{p}=(a+1)\hat{i}+a\hat{j}+a\hat{k},\overrightarrow{q}=a\hat{i}+(a+1)\hat{j}+a\hat{k}$ and $\overrightarrow{r}=a\hat{i}+a\hat{j}+(a+1)\hat{k},(a\in R)$ are coplanar and $3(\overrightarrow{p}.\overrightarrow{q}{)}^2-\lambda |\overrightarrow{r}\times \overrightarrow{q}{|}^2=0,$ then value of` $\lambda$ is

Answer 1

As $\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}$ are coplanar,
$\left|\begin{array}{ccc}a+1& a& a\\ a& a+1& a\\ a& a& a+1\end{array}\right|=0$

$R_1\to R_1+R_2+R_3$

$\Rightarrow \left|\begin{array}{ccc}3a+1& 3a+1& 3a+1\\ a& a+1& a\\ a& a& a+1\end{array}\right|=0$

$\Rightarrow (3a+1)\left|\begin{array}{ccc}1& 1& 1\\ a& a+1& a\\ a& a& a+1\end{array}\right|=0$

$C_2\to C_2-C_1,C_3\to C_3-C_1$

$\Rightarrow (3a+1)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 0\\ a& 0& 1\end{array}\right|=0$
$\Rightarrow 3a+1=0\Rightarrow a=-\frac{1}{3}$

$\overrightarrow{p}=\frac{1}{3}(2\hat{i}-\hat{j}-\hat{k}),\overrightarrow{q}=\frac{1}{3}(-\hat{i}+2\hat{j}-\hat{k})$,
$\overrightarrow{r}=\frac{1}{3}(-\hat{i}-\hat{j}+2\hat{k})$

$\overrightarrow{r}\times \overrightarrow{q}=\frac{1}{9}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -1& 2\\ -1& 2& -1\end{array}\right|$

$\overrightarrow{r}\times \overrightarrow{q}=\frac{1}{9}(-3\hat{i}-3\hat{j}-3\hat{k})$
$=-\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

$|\overrightarrow{r}\times \overrightarrow{q}{|}^2=\frac{1}{3}$
$\overrightarrow{p}.\overrightarrow{q}=\frac{1}{9}(-2-2+1)=-\frac{1}{3}$

$3(\overrightarrow{p}.\overrightarrow{q}{)}^2-\lambda |\overrightarrow{r}\times \overrightarrow{q}{|}^2=0\Rightarrow \frac{1}{3}-\lambda \times \frac{1}{3}=0$
$\Rightarrow \lambda =1$