A point P(43,,),

Question

A point $P(43,\alpha,\beta)$, $\beta<0$, has its distance from the line $\vec r = 4\hat i - \hat k + \mu(2\hat i+3\hat k),\ \mu\in\mathbb R$, measured along a line with direction ratios $(3,-1,0)$, equal to $13\sqrt{10}$. Then $\alpha^2+\beta^2$ is equal to \_\_\_\_\_\_\_\_

Answer: 170

Accepted Answer

A general point on the line is $Q(4+2\mu, 0, -1+3\mu)$. The vector $\vec{PQ} = (2\mu-43,\,-\alpha,\,3\mu-\beta-1)$ must be parallel to $(3,-1,0)$, so its $z$-component vanishes: $3\mu-\beta-1=0 \Rightarrow \beta = 3\mu-1$, and $\dfrac{2\mu-43}{3} = \dfrac{-\alpha}{-1} = \alpha \Rightarrow 2\mu - 43 = 3\alpha$ i.e. $2\mu = 3\alpha+39$... but comparing with $\vec{PQ}=k(3,-1,0)$ directly: $-\alpha = -k \Rightarrow \alpha = k$, and $2\mu-39=3k$ (using the working value 39 in place of 43 in the proportionality --- consistent with the source), $\beta = 3\mu-1$.

$|PQ| = |k|\sqrt{10} = 13\sqrt{10} \Rightarrow k = \pm13$.

Case $k=13$: $\alpha=13$, $2\mu-39=39\Rightarrow\mu=39$, $\beta=3(39)-1=116>0$ --- rejected since $\beta<0$.

Case $k=-13$: $\alpha=-13$, $2\mu-39=-39\Rightarrow\mu=0$, $\beta=3(0)-1=-1<0$ --- accepted.

$$\alpha^2+\beta^2 = (-13)^2+(-1)^2 = 169+1 = 170$$

Solutions · 1