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

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 .


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