Let P_1: r(2 i+ j-3 k)=4 be a plane.

Question

Let $P_1:\ \vec r\cdot(2\hat i+\hat j-3\hat k)=4$ be a plane. Let $P_2$ be another plane which passes through the points $(2,-3,2),\ (2,-2,-3),\ (1,-4,2)$. If the direction ratios of the line of intersection of $P_1$ and $P_2$ are $16,\alpha,\beta$, then the value of $\alpha+\beta$ is equal to:

Answer: 28

Accepted Answer

Normal to $P_1$: $(2,1,-3)$.

Vectors in $P_2$: $\vec{AB}=(0,1,-5)$, $\vec{BC}=(-1,-2,5)$; normal $\vec{AB}\times\vec{BC}=(-5,5,1)$.

Direction of the intersection line: $(2,1,-3)\times(-5,5,1)=(16,13,15)$.
$$\alpha=13,\ \beta=15 \implies \alpha+\beta=28.$$

Solutions · 1