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Angle bisector in a triangle: divides the opposite side in the ratio of the sides, or points along the sum of unit vectors

This type of question, asked 5 times: 27 Jan 2024, 29 Jan 2024, 28 Jan 2025, 28 Jan 2026 S1, 28 Jan 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    27 Jan 2024

    166
    The position vectors of the vertices A,BA,B and CC of a triangle are 2i^−3j^+3k^,2i^+2j^+3k^2\widehat{i}- 3\widehat{j}+ 3\widehat{k},2\widehat{i}+ 2\widehat{j}+ 3\widehat{k} and −i^+j^+3k^-\widehat{i}+\widehat{j}+ 3\widehat{k} respectively. Let ll denotes the length of the angle bisector ADAD of ∠BAC\angle BAC where DD is on the line segment BCBC, then 2l22l^{2} equals:

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  2. Question 2

    29 Jan 2024

    73
    Let OO be the origin and the position vector of AA and BB be 2i^+2j^+k^2\widehat{i}+ 2\widehat{j}+\widehat{k} and 2i^+4j^+4k^2\widehat{i}+ 4\widehat{j}+ 4\widehat{k} respectively. If the internal bisector of ∠AOB\angle AOB meets the line ABAB at CC, then the length of OCOC is

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  3. Question 3

    28 Jan 2025

    127
    Let A, B, C be three points in xy-plane, whose position vector are given by 3i^+j^,i^+3j^\sqrt{3}\widehat{i}+\widehat{j},\widehat{i}+\sqrt{3}\widehat{j} and a i^+(1−a)j^\widehat{i}+ (1 - a)\widehat{j} respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors OA→\overrightarrow{OA} and OB→\overrightarrow{OB} is 92\frac{9}{\sqrt{2}}, then the sum of all the possible values of aa is:

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  4. Question 4

    28 Jan 2026 S1

    75
    Let PQRPQR be a triangle such that PQ→=−2i^−j^+2k^\overrightarrow{PQ} = - 2\widehat{i} - \widehat{j} + 2\widehat{k} and PR→=ai^+bj^−4k^,a,b∈Z\overrightarrow{PR} = a\widehat{i} + b\widehat{j} - 4\widehat{k},a,b \in \mathbb{Z}. Let SS be the point on QR, which is equidistant from the lines PQ and PR. If ∣PR→∣=9|\overrightarrow{PR}| = 9 and PS→=i^−7j^+2k^\overrightarrow{PS} = \widehat{i} - 7\widehat{j} + 2\widehat{k}, then the value of 3a−4b3a - 4b is ____\_\_\_\_ .

  5. Question 5

    28 Jan 2026 S2

    67
    Let P be a point in the plane of the vector AB→=3i^+j^−k^\overrightarrow{AB}= 3\widehat{i}+\widehat{j}-\widehat{k} and AC→=i^−j^+3k^\overrightarrow{AC}=\widehat{i}-\widehat{j}+ 3\widehat{k} such that PP is equidistant from the lines AB and AC . If ∣AP→∣=52|\overrightarrow{AP}| =\frac{\sqrt{5}}{2}, then the area of the triangle ABPABP is :

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