Question

The vectors □(→┬a ) and □(→┬b ) are non-collinear. Find for what value of x, the vectors □(→┬c )=(x-2) □(→┬a )+□(→┬b ) and □(→┬d ) =(2x+1) □(→┬a )- □(→┬b ) are collinear.

Answer 1

Answer on Question #46178 – Math - Vector Calculus

Problem.

The vectors $\square (\rightarrow_{\mathsf{T}} a)$ and $\square (\rightarrow_{\mathsf{T}} b)$ are non-collinear. Find for what value of $x$, the vectors $\square (\rightarrow_{\mathsf{T}} c) = (x - 2)\square (\rightarrow_{\mathsf{T}} a) + \square (\rightarrow_{\mathsf{T}} b)$ and $\square (\rightarrow_{\mathsf{T}} d) = (2x + 1)\square (\rightarrow_{\mathsf{T}} a) - \square (\rightarrow_{\mathsf{T}} b)$ are collinear.

Solution.

The vectors $\vec{c} = (x - 2)\vec{a} + \vec{b}$ and $\vec{d} = (2x + 1)\vec{a} - \vec{b}$ are colliner if there exists $\lambda$ such that $\vec{c} = \lambda \vec{d}$.

Then $(x - 2)\vec{a} + \vec{b} = \lambda((2x + 1)\vec{a} - \vec{b})$. Hence $\big((x - 2) - \lambda(2x + 1)\big)\vec{a} = (-1 - \lambda)\vec{b}$. The vectors $\vec{a}$ and $\vec{b}$ are non-collinear, so $(x - 2) - \lambda(2x + 1) = 0$ and $-1 - \lambda = 0$. Hence $\lambda = -1$ and $(x - 2) + (2x + 1) = 0$, $3x = 1$.

Therefore $x = \frac{1}{3}$.

Answer. $x = \frac{1}{3}$.

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