Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line (x + 1)/2 = (y - 3)/3 = (z - 1)/(-1).

Question

Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \(\cfrac{\text x+1}2=\cfrac{y-3}3=\cfrac{z-1}{-1}.\) Hence or otherwise deduce the length of the perpendicular.

Answer 1

Let point P = (5, 4, 2) and Q be the foot of the perpendicular drawn from to P the line \(\cfrac{\text x+1}2=\cfrac{y-3}3=\cfrac{z-1}{-1}.\). Q is a point on the given line. So, for some α, Q is given by

\(\cfrac{\text x+1}2=\cfrac{y-3}3=\cfrac{z-1}{-1}\) = α (say)

⇒ x = 2α – 1, y = 3α + 3, z = –α + 1

Thus, Q = (2α – 1, 3α + 3, –α + 1)

Now, we find the direction ratios of PQ.

Recall the direction ratios of a line joining two points (x₁, y₁, z₁) and (x₂, y₂, z₂) are given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

Here, (x₁, y₁, z₁) = (5, 4, 2) and (x₂, y₂, z₂) = (2α – 1, 3α + 3, –α + 1)

⇒ Direction Ratios of PQ are ((2α – 1) – (5), (3α + 3) – (4), (–α + 1) – (2))

⇒ Direction Ratios of PQ are (2α – 6, 3α – 1, –α – 1)

PQ is perpendicular to the given line, whose direction ratios are (2, 3, –1).

We know that if two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular to each other, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

⇒ (2)(2α – 6) + (3)(3α – 1) + (–1)(–α – 1) = 0

⇒ 4α – 12 + 9α – 3 + α + 1 = 0

⇒ 14α – 14 = 0

⇒ 14α = 14

∴ α = 1

We have Q = (2α – 1, 3α + 3, –α + 1)

⇒ Q = (2×1 – 1, 3×1 + 3, –1 + 1)

∴ Q = (1, 6, 0)

Using the distance formula, we have

Thus, the required foot of perpendicular is (1, 6, 0) and the length of the perpendicular is \(2\sqrt6\) units.