InterviewSolution
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InterviewSolution
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Let `vecAB = 3 hati + hatj - hatk and vecAC = hati -hatj + 3hatk` and a point P on the line segment BC is equidistant from AB and AC, then `vec (AP)` isA. `2 hati - hatk `B. `hati -2 hatk`C. `2hati +hatk`D. none of these |
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Answer» Correct Answer - C Clearly, a point equidistant from AB and AC is on the bisector of the `angle BAC.` A vector along the bisector of `angle BAC` is `(vec(AB))/(|vec(AB)|)=(vec(AC))/(|vec(AC)|)= (1)/(sqrt(11))(4hati +2hatk)=(2)/(sqrt(11))(2hati+hatk)` Let ` vec(AP) = lambda(2hati + hatk )` `therefore vec(BP) = vec(AP)-vec(AB) =lambda(2 hati+hat k)-(3hati + hatj -hatk)` `rArr vec(BP) = (2lambda -3) hati - hat j + (lambda + 1) hat k` Also, `vec(BC) = vec(AC) - vec(AB) = - 2 hati - 2 hatj + 4 hatk` Since ` vec(BP) ||vec(BC).` Therefore, `vec(BP) = t vec(BC)` `rArr (2lambda -3) hati -hatj + (lambda +1) hatk = t(-2hati - 2 hatj + 4hatk)` `rArr 2 lambda -3= -2t, -1= -2t, lambda +1= 4t` `rArr lambda =1, t=(1)/(2)` `therefore vec(AP) = 2 hati + hatk`. |
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