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Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.3 Question 5 Maths Textbook Solution.

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Answer:\begin{aligned} &\frac{3 \vec{a}+5 \vec{c}}{8}=\frac{2 \vec{b}+6 \vec{d}}{8} \\ \end{aligned}

Hint: Use vector algebra

Given: \begin{aligned} &\qquad 3 \vec{a}-2 \vec{b}+5 \vec{c}-6 \vec{d}=0 \\ \end{aligned}

Solution: \qquad 3\vec{a} +5 \vec{c}=2 \vec{b}+6\vec{d}

  \begin{aligned} &\frac{3 \vec{a}+5 \vec{c}}{8}=\frac{2 \vec{b}+6 \vec{d}}{8} \end{aligned}

 

AC and BD are intersect at the point.

A,B,C,D are co- planner.

P.V of point of intersection of line segment AC and BD are   \begin{aligned} &\frac{3 \vec{a}+5 \vec{c}}{8}\: or\: \frac{2 \vec{b}+6 \vec{d}}{8} \\ \end{aligned}

 

 

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