If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between the vectors and ]

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Solution

The given vertices of triangle are A(1,2,3), B(-1,0,0) and C(0,1,2). Also ∠ABC is the angle between BA → and BC → .

Let, ∠ABCbe given by θ,

The angle between vector BA → and BC → is given by,

cosθ= BA → BC → | BA || BC | (1)

BA → =( 1−(−1) ) i ^ +( 2−0 ) j ^ +(3−0) k ^ =2 i ^ +2 j ^ +3 k ^

| AB | → = 2 2 + 2 2 + 3 2 = 17

BC → =( 0−( −1 ) ) i ^ +( 1−0 ) j ^ +(2−0) k ^ = i ^ + j ^ +2 k ^

| BC | → = 1 2 + 1 2 + (2) 2 = 1+1+4 = 6

BA. → BC → =( 2 i ^ +2 j ^ +3 k ^ )( i ^ + j ^ +2 k ^ ) =2×1+2×1+3×2 =10

Substitute the values of, BA. → BC → , | AB | → and | BC → |in (1),

cosθ= BA → BC → | BA → | | BC | → cosθ= 10 17 6 θ= cos −1 10 102

Thus, the angle between BA → and BC → of triangle with vertices A(1,2,3), B(-1,0,0) and C(0,1,2) is θ= cos −1 10 102 .

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