MHT CET · Maths · Vector Algebra
The incenter and centroid of the triangle, whose vertices are \(A \equiv(0,3,0), B \equiv(0,0,4)\), and \(C \equiv(0,3,4)\), are respectively given by
- A \((0,-2,-3),\left(0,-2, \frac{8}{3}\right)\)
- B \((0,-2,3),\left(0,2,-\frac{8}{3}\right)\)
- C \(\left(0,2, \frac{8}{3}\right),(0,2,3)\)
- D \((0,2,3),\left(0,2, \frac{8}{3}\right)\)
Answer & Solution
Correct Answer
(D) \((0,2,3),\left(0,2, \frac{8}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(A \equiv(0,3,0), B \equiv(0,0,4)\), and \(C \equiv(0,3,4)\)
\(a=B C=3\)
\(b=C A=4\)
\(c=A B=5\).

\(\text {Now, Incentre } \equiv(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c},\) \( \frac{a z_1+b z_2+c z_3}{a+b+c}) \)
\( \equiv(\frac{3 \times 0+4 \times 0+5 \times 0}{3+4+5}, \frac{3 \times 3+4 \times 0+5 \times 3}{3+4+5},\) \(\frac{3 \times 0+4 \times 4+5 \times 4}{3+4+5}) \)
\( \equiv(0,2,3) \)
\( \text {and centroid } \equiv(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3},\) \(\frac{z_1+z_2+z_3}{3})\)
\(\begin{aligned} & \equiv\left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) \\ & \equiv\left(0,2, \frac{8}{3}\right)\end{aligned}\)
\(a=B C=3\)
\(b=C A=4\)
\(c=A B=5\).

\(\text {Now, Incentre } \equiv(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c},\) \( \frac{a z_1+b z_2+c z_3}{a+b+c}) \)
\( \equiv(\frac{3 \times 0+4 \times 0+5 \times 0}{3+4+5}, \frac{3 \times 3+4 \times 0+5 \times 3}{3+4+5},\) \(\frac{3 \times 0+4 \times 4+5 \times 4}{3+4+5}) \)
\( \equiv(0,2,3) \)
\( \text {and centroid } \equiv(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3},\) \(\frac{z_1+z_2+z_3}{3})\)
\(\begin{aligned} & \equiv\left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) \\ & \equiv\left(0,2, \frac{8}{3}\right)\end{aligned}\)
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