If B is the image of A(,2,1) in the line x-23=y-12=z1, and B=(2+1, ^2-3, -12), then the po…

Question

If $B$ is the image of $A(\alpha,2\alpha,1)$ in the line $\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}$, and $B=\left(2\alpha+1,\ \alpha^2-3\alpha,\ \dfrac{\alpha-1}{2}\right)$, then the possible value(s) of $\alpha$ is (are):

Accepted Answer

The midpoint of $AB$ must lie on the line:
$$M=\left(\frac{3\alpha+1}{2},\ \frac{\alpha^2-\alpha}{2},\ \frac{\alpha+1}{4}\right).$$
Substituting into $\dfrac{x-2}{3}=\dfrac{y-1}{2}=z$:
$$\frac{\alpha-1}{2}=\frac{\alpha+1}{4}=\frac{\alpha^2-\alpha-2}{4}.$$
From the first equality: $2(\alpha-1)=\alpha+1 \implies \alpha=3$. This value also satisfies the third ratio, and $\vec{AB}=(4,-6,0)$ is indeed perpendicular to the line's direction $(3,2,1)$ at $\alpha=3$ ($12-12+0=0$).

So the only possible value is $\alpha=3$.

Solutions · 1