Question

Find the foot of the perpendicular on the z axis from the coordinates (1 , -2, 6).

3d geometry. Please explain with a diagram if possible.

Answer 1

$\large\underline{\sf{Solution-}}$

We know, z - axis passes through the point (0, 0, 0) and having direction ratios (0, 0, 1). So, equation of z - axis is given by

$\rm \: \dfrac{x - 0}{0} = \dfrac{y - 0}{0} = \dfrac{z - 0}{1} = k$

$\rm \: \dfrac{x}{0} = \dfrac{y}{0} = \dfrac{z}{1} = k$

So, any point on z - axis, say P (0, 0, k) ----(1)

Now, Q(1, - 2 6) be any point from where perpendicular draw on z - axis meet at the point P.

So, direction ratios of PQ = (1 - 0, - 2 - 0, 6 - k) = (1, - 2, 6 - k)

Now, PQ is perpendicular to z - axis.

It means,

$\rm \: 1 \times 0 - 2 \times 0 + 1 \times (6 - k) = 0$

$\rm \: 0 + 0 + (6 - k) = 0$

$\rm \: 6 - k = 0$

$\rm\implies \:k \: = \: 6$

So, on substituting the value of k in equation (1), we get

$\rm \: Coordinates \: of \: foot \: of \: perpendicular = (0,0,6)$

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Short Cut Trick :-

If P( a, b, c) be any point, then coordinates of foot of perpendicular drawn from point P on z - axis be (0, 0, c)

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Formulae Used :-

1. Equation of line which passes through the point (a, b, c) and having direction ratios (p, q, r) is given by

$\rm \: \dfrac{x - a}{p} = \dfrac{y - b}{q} = \dfrac{z - c}{r} = k$

2. Two lines having direction ratios (a, b, c) and (d, e, f) are perpendicular iff ad + be + cf = 0

Answer 2

EXPLANATION.

To find foot of the perpendicular on the z- axis.

Co-ordinates = (1, - 2, 6).

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.

Point on z - axis .

It means x and y both are 0.

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

Directions cosines vectors are : (1, - 2, z - 6).

As we know that,

Dot product of perpendicular is always equal to zero.

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

$\implies \vec{a_{1}} = \vec{i} -2\vec{j} + (z - 6)\vec{k}$

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

$\implies [\vec{i} -2\vec{j} + (z - 6)\vec{k} ]. (z \vec{k}) = 0$

$\implies z - 6 = 0$

$\implies z = 6$

Co-ordinates of foot of perpendicular on the z - axis : (0, 0, 6).