Q34 / 50 · Manipal (MET) 2023 · Mathematics
If the vectors $p=(a+1) \hat{\mathbf{i}}+a \hat{\mathbf{j}}+a \hat{\mathbf{k}}$, $q=a \hat{\mathbf{i}}+(a+1) \hat{\mathbf{j}}+a \hat{\mathbf{k}}$ and $r=a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+(a+1) \hat{\mathbf{k}}$ $(a \in R)$ are coplanar, then the value of $a$ is____
Correct answer: -0.33
Solution
Given, vectors are
$\begin{aligned} & \mathbf{p}=(a+1) \hat{\mathbf{i}}+a \hat{\mathbf{j}}+a \hat{\mathbf{k}} \\ & \mathbf{q}=a \hat{\mathbf{i}}+(a+1) \hat{\mathbf{j}}+a \hat{\mathbf{k}}\end{aligned}$
and $\mathbf{r}=a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+(a+1) \hat{\mathbf{k}}$
$[p q r]=0$
$\begin{aligned} & \Rightarrow\left|\begin{array}{ccc}a+1 & a & a \\ a & a+1 & a \\ a & a & a+1\end{array}\right|=0 \\ & \Rightarrow(a+1)\left[(a+1)^2-a^2\right]-a\left[a(a+1)-a^2\right] \\ & \quad+a\left[a^2-a(a+1)\right]=0\end{aligned}$
$\begin{aligned} & \Rightarrow(a+1)[2 a+1]-2 a[a]=0 \\ & \Rightarrow 2 a^2+3 a+1-2 a^2=0 \\ & \Rightarrow a=-\frac{1}{3}=-0.33\end{aligned}$
Hence, the correct answer is -0.33.