Question

Find the perpendicular distance of a point (1, -3, -7) from z axis.

3d geometry. Please explain with diagram if possible.

Answer 1

EXPLANATION.

Perpendicular distance of a point (1, - 3, - 7) from z - axis.

As we know that,

Concept of :

Distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂).

d = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)².

Using this concept on the equation, we get.

Point on z - axis.

It means x and y both will zero.

Co-ordinates on z - axis : (0, 0, z).

Directions cosines vectors are : (1, - 3, z + 7).

As we know that,

Dot product of perpendicular is always equal to zero.

$\implies \vec{a_{1}}. \vec{a_{2}} = 0.$

$\implies \vec{a_{1}} = \vec{i} - 3\vec{j} + (z + 7) \vec{k}$

$\implies \vec{a_{2}} = z \vec{k}$

$\implies [\vec{i} - 3\vec{j} + (z + 7) \vec{k}] . [z \vec{k} ] = 0$

$\implies z + 7 = 0.$

$\implies z = - 7.$

Co-ordinates of foot of perpendicular on the z - axis : (0, 0, - 7).

                                                                                                               

MORE INFORMATION.

Let $\vec{a} = \vec{a_{1}} \hat{i} + \vec{a_{2}} \hat{j} + \vec{a_3}} \hat{k}$ the angles which this vector makes with the + ve directions are called Direction Angles and their Cosines are called direction cosines.

$cos(\alpha ) = \dfrac{a_{1}}{|\vec{a}|} \ \ \ cos(\beta ) = \dfrac{a_{2}}{|\vec{a}|} \ \ \ cos(\gamma) = \dfrac{a_{3}}{|\vec{a}|}$

$\implies cos^{2} (\alpha ) + cos^{2} (\beta ) + cos^{2} (\gamma) = 1$