Let a=(1+t) i+(1-t) j+ k, b=(1-t) i+(1+t) j+2 k, c=t i-t j+ k, t R, be such that a+ b+ c= …

Question

Let $\vec a=(1+t)\hat i+(1-t)\hat j+\hat k$, $\vec b=(1-t)\hat i+(1+t)\hat j+2\hat k$, $\vec c=t\hat i-t\hat j+\hat k$, $t\in\mathbb R$, be such that $\alpha\vec a+\beta\vec b+\gamma\vec c=\vec 0 \Rightarrow \alpha=\beta=\gamma=0$. The set of all such $t$ is:

Accepted Answer

The condition holds iff $\vec a,\vec b,\vec c$ are linearly independent, i.e. $[\vec a\ \vec b\ \vec c]\ne0$:
$$\begin{vmatrix}1+t&1-t&1\1-t&1+t&2\t&-t&1\end{vmatrix}\ne0$$

Expanding: $(1+t)(1+t+2t)-(1-t)(1-t-2t)+(t^2-t-t-t^2) \ne 0 \implies (3t^2+4t+1)-(3t^2-4t+1)-2t\ne0 \implies 6t\ne0$

So $t\ne0$, giving $t\in\mathbb R-\{0\}$.

Solutions · 1