# If the vectors,

Question:

If the vectors, $\vec{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$

$\vec{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$ and $\vec{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k} \quad(a \in \mathrm{R})$ are complanar and $3(\vec{p} \cdot \vec{q})^{2}-\lambda|\vec{r} \times \vec{q}|^{2}=0$, then the value of $\lambda$ is_________.

Solution:

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

$\Rightarrow 3 a+1=0 \Rightarrow a=-\frac{1}{3}$

The given vectors

$\vec{p}=\frac{2}{3} \hat{i}-\frac{1}{3} \hat{j}-\frac{1}{3} \hat{k}=\frac{1}{3}(2 \hat{i}-\hat{j}-\hat{k})$

$\vec{q}=\frac{1}{3}(-\hat{i}+2 \hat{j}-\hat{k})$

$\vec{r}=\frac{1}{3}(-\hat{i}-\hat{j}+2 \hat{k})$

Now, $\vec{p} \cdot \vec{q}=\frac{1}{9}(-2-2+1)=-\frac{1}{3}$

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

$=\frac{1}{9}(i(4-1)-j(-2-1)+k(1+2))$

$=\frac{1}{9}(3 i+3 j+3 k)=\frac{i+j+k}{3}$

$|\vec{r} \times \vec{q}|=\frac{1}{3} \sqrt{3} \Rightarrow|\vec{r} \times \vec{q}|^{2}=\frac{1}{3}$

$3(\vec{p} \cdot \vec{q})^{2}-\lambda|\vec{r} \times \vec{q}|^{2}=0$

$\Rightarrow \quad 3 \cdot \frac{1}{9}-\lambda \cdot \frac{1}{3}=0 \Rightarrow \lambda=1$