Question

If the vectors $(p+1)\hat{i}-3\hat{j}+p\hat{k},p\hat{i}+(p+1)\hat{j}-3\hat{k},-3\hat{i}+p\hat{j}+(p+1)\hat{k}$ are linearly dependent, then find the real value of $p$.

Answer 1

The given vectors are linearly dependent.

Hence, $\left|\begin{array}{ccc}p+1& -3& p\\ p& p+1& -3\\ -3& p& p+1\end{array}\right|=0$

This is a circulant matrix. Hence, by circular Matrix Property

if $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

Then by Expansion of determinant

$a^3+b^3+c^3-3abc=0$

Or

$\frac{1}{2}(a+b+c)\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]$

$⟹a+b+c=0Ora=b=c$

So Either, $p+p+1-3=0$ or $p=p+1=-3$.
The second case is not possible.Because for $p=-3orp=-4$ a=b=c not satisfying

Hence, $2p-2=0$

Hence, $p=1$