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The position vector of A and B are 2i+2j+k and 2i+4j+4k the length of the internal bisector of angle BOA of triangle AOB is

The position vector of A and B are 2i+2j+k and 2i+4j+4k the length of the internal bisector of angle BOA of triangle AOB is

Grade:12

1 Answers

Navendu Shekhar
13 Points
5 years ago
Answer is (96/9)^(1/2)Steps:1. First find the unit vector in the direction of A and B.2. Now you can have the unit vector in the direction of internal angle bisector by the formula (unit vectoA+unit vector B) /2. That is [3t 4t 3t].3. Now you also have the position vector AB, since you have the postion vector of A and B. Suppose internal bisector cuts at point C of the line AB.4. Suppose C divides line AB in the ratio k:1 then the point C will be (kB+A) /(k+1). This way you have the position vector of internal bisector. That is (2k+2,4k+2,4k+1)/(k+1).5. Now equate both the vectors since they reprent the same vector.6. This gives k=1/2.7. Solve it by substituting the value of k to get the length of internal bisector that is (96/9)^(1/2).

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