Let a line pass through the origin and be perpendicular to the lines _1: r=( i-11 j-7 k)+(…

Question

Let a line $\ell$ pass through the origin and be perpendicular to the lines $\ell_1:\ \vec r=(\hat i-11\hat j-7\hat k)+\lambda(\hat i+2\hat j+3\hat k)$, $\lambda\in\mathbb R$, and $\ell_2:\ \vec r=(-\hat i+\hat k)+\mu(2\hat i+2\hat j+\hat k)$, $\mu\in\mathbb R$. If $P$ is the point of intersection of $\ell$ and $\ell_1$, and $Q(\alpha,\beta,\gamma)$ is the foot of perpendicular from $P$ on $\ell_2$, then $9(\alpha+\beta+\gamma)$ is equal to:

Answer: 5

Accepted Answer

Direction of $\ell$: $(1,2,3)\times(2,2,1)=(-4,-5,-2)$ (up to sign).

$\ell:\ \gamma_0(-4,5,-2)$ (parametrized). Intersection with $\ell_1$: solving $-4\gamma_0=1+\lambda,\ 5\gamma_0=-11+2\lambda,\ -2\gamma_0=-7+3\lambda$ gives $\gamma_0=-1$, so $P=(4,-5,2)$.

Foot of perpendicular $Q$ on $\ell_2$: $Q(-1+2\mu,\ 2\mu,\ 1+\mu)$; $\vec{PQ}\cdot(2,2,1)=0$:
$$-2+4\mu+4\mu+1+\mu=0 \implies \mu=\frac19.$$
$$Q=\left(-\frac79,\frac29,\frac{10}{9}\right).$$
$$9(\alpha+\beta+\gamma)=9\left(-\frac79+\frac29+\frac{10}{9}\right)=5.$$

Solutions · 1