MHT CET · Maths · Vector Algebra
The vertices of triangle \(\mathrm{ABC}\) are \(\mathrm{A} \equiv(3,0,0) ; \mathrm{B} \equiv(0,0,4)\); \(\mathrm{C} \equiv(0,5,4)\). Find the position vector of the point in which the bisector of angle A meets \(\mathrm{BC}\) is
- A \(5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}\)
- B \(\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{k}}}{3}\)
- C \(\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}}{13}\)
- D \(\frac{5 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{k}}}{3}\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{AD}\) be the angle bisector of angle \(\mathrm{A}\) which divides \(\mathrm{BC}\) in the ratio
\(\mathrm{AB}: \mathrm{AC}\)
Here \(\mathrm{AB}=\sqrt{9+16}=\sqrt{25}\) and
\(
\begin{aligned}
& \mathrm{AC}=\sqrt{9+25+16} \\
& =\sqrt{50}
\end{aligned}
\)
\(\therefore \mathrm{D}\) divides \(\mathrm{BC}\) in the ratio \(\sqrt{25}: \sqrt{50}\) i.e., \(1: 2\)
\(\therefore\) Position vector of \(\mathrm{D}=\frac{(4)(2) \hat{\mathrm{k}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{1+2}=\frac{5 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{3}\)
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