Find the directional cosines of the sides of the triangle whose vertices are (3,5,–4), (–1,1,2), (–5,–5,–2).

Question

Philoid · Accepted Answer

Let us write the given points as:

⇒ A = (3,5,–4)

⇒ B = (–1,1,2)

⇒ C = (–5,–5,–2)

Let us assume the direction ratios of sides AB be (r₁,r₂,r₃), BC be (r₄,r₅,r₆) and CA be (r₇,r₈,r₉)

We know that direction ratios for a line passing through points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (x₂–x₁, y₂–y₁, z₂–z₁).

Let us find the direction ratios for the side AB

⇒ (r₁,r₂,r₃) = (–1–3, 1–5, 2–(–4))

⇒ (r₁,r₂,r₃) = (–1–3, 1–5, 2+4)

⇒ (r₁,r₂,r₃) = (–4,–4,6)

Let us find the direction ratios for the side BC

⇒ (r₄,r₅,r₆) = (–5–(–1), –5–1, –2–2)

⇒ (r₄,r₅,r₆) = (–5+1, –5–1, –2–2)

⇒ (r₄,r₅,r₆) = (–4,–6,–4)

Let us find the direction ratios for the side CA

⇒ (r₇,r₈,r₉) = (3–(–5), 5–(–5), –4–(–2))

⇒ (r₇,r₈,r₉) = (3+5, 5+5, –4+2)

⇒ (r₇,r₈,r₉) = (8,10,–2)

Let us assume be the direction cosines of line AB, be the direction cosines of line BC and be the direction cosines of line CA.

We know that for a line of direction ratios r₁, r₂, r₃ and having direction cosines has the following property.

⇒

Let us follow the above property and find the direction cosines of each side.

Now, let’s find the direction cosines of side AB,

⇒

The direction cosines for the side AB is .

Let’s find the directional cosines for the side BC,

⇒

The direction cosines for the sides BC is .

Let’s find the direction cosines for the side CA,

⇒

The direction cosines for the sides CA is .