alleryIf $A(0, 3, 0), B(0, 0, 4), C(0, 3, 4)$ are vertices of a triangle, the position vector of its **incentre** is:
A$2\hat j + 3\hat k$
B$3\hat j + 2\hat k$
C$\hat j + \hat k$
D$(\hat j + \hat k)/2$
Answer & Solution
Correct answer: A. $2\hat j + 3\hat k$
**Compute side lengths:** $|AB| = 5, |AC| = 4, |BC| = 3$.
Incentre $H = (|BC|\,\vec a + |AC|\,\vec b + |AB|\,\vec c)/(|BC|+|AC|+|AB|)$.
**Substituting:** $H = (3(3\hat j) + 4(4\hat k) + 5(3\hat j + 4\hat k))/12 = (24\hat j + 36\hat k)/12 = 2\hat j + 3\hat k$, i.e. the point $(0, 2, 3)$.
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Related questions
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