Question

Using projection show that the line passing through (-1, 8, 8) and (6, 2, 0) is perpendicular to the line passing through (4, 2, 3) and (2, 1, 2).

Answer 1

Answer on Question #45096 – Math - Analytic Geometry

Problem.

Using projection show that the line passing through $(-1, 8, 8)$ and $(6, 2, 0)$ is perpendicular to the line passing through $(4, 2, 3)$ and $(2, 1, 2)$.

Solution.

The direction vector of the first line is $\vec{a}(7, -6, -8)$ and the direction vector of the second line is $\vec{b}(-2, -1, -1)$. The projection of the vector $\vec{a}$ on the vector $\vec{b}$ is equal to

$$\overrightarrow{a_0} = \frac{(\vec{a}, \vec{b})}{|\vec{b}|^2} \vec{b},$$

where $(\vec{a}, \vec{b})$ is inner product of $(\vec{a}, \vec{b})$.

$$\overrightarrow{a_0} = \frac{7 \cdot (-2) + (-6) \cdot (-1) + (-8) \cdot (-1)}{(-2)^2 + (-1)^2 + (-1)^2} \vec{b} = \vec{0},$$

so the projection of the vector $\vec{a}$ on the vector $\vec{b}$ is equal to $\vec{0}$. Hence the vector $\vec{a}$ is perpendicular to the vector $\vec{b}$ or the line passing through $(-1, 8, 8)$ and $(6, 2, 0)$ is perpendicular to the line passing through $(4, 2, 3)$ and $(2, 1, 2)$ ($\vec{a}$ and $\vec{b}$ are direction vectors of this lines).

www.AssignmentExpert.com