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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors Exercise 22.3 Question 7 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.3 Question 7 Maths Textbook Solution.

Answers (1)

Answer: – we need to prove that  I=\frac{\alpha \vec{a}+(\beta+\gamma) \vec{d}}{\alpha+\beta+\gamma}

Hint: Use vector algebra.

Given: The vertices A,B,C of triangle ABC have respectively. P.V  \vec{a},\vec{b},\vec{c} with respect to the given origin 0.

Solution : 

               

Let the PV of A,B and C w.r.t some origin 0 be \vec{a},\vec{b},\vec{c}  respectively.  let DB be a point on BC where  bisector of angle D meets.

 let \vec{d}  be P.V of D which divide CB internally in the ratio β and \gamma where \beta =\mid \vec{AC}\mid and \gamma =\mid \vec{AB}\mid

Thus \beta =\mid \vec{c}-\vec{a}\mid and  \gamma =\mid \vec{b}-\vec{a}\midby section formula, the P.V of D is given by

\begin{aligned} &=\overrightarrow{O D}=\frac{\beta \vec{b}+\gamma \vec{c}}{\beta+\gamma} \\ &=\alpha=|\vec{b}-\vec{c}| \end{aligned}

 

In centre is the concurrent point angle bisector and in centre divides the line AD in the ratio

=a=\beta -\gamma

So the P.V of in centre is given by I=\frac{\alpha \vec{a}+(\beta+\gamma) \vec{d}}{\alpha+\beta+\gamma}

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