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  • Need solution for RD Sharma maths class 12 chapter Scaler and Dot Product exercise 23.2 question 3

Need solution for RD Sharma maths class 12 chapter Scaler and Dot Product exercise 23.2 question 3

Answers (1)

Answer: (A C)^{2}=(A B)^{2}+(B C)^{2}

Hint:  Pythagoras theorem

Given: P.T by vector method in right angle triangle the square of hypotenuse

is equal to sum of other two sides.

Solution: Let ABC is right angled triangle at B.

The side opposite to right angle is hypotenuse

Hence, (A C)^{2}=(A B)^{2}+(B C)^{2}

Hence, \vec{a} \cdot \vec{b}=0 \quad \rightarrow(1)

By triangle law of vector addition

A C=\vec{b}-\vec{a}

Squaring both sides,

\begin{aligned} &(A C)^{2}=(\vec{b}-\vec{a})^{2} \\\\ &(A C)^{2}=(\vec{a})^{2}+(\vec{b})^{2}-2(\vec{a} \cdot \vec{b}) \end{aligned}

Use eqn(1) we get                                    Now since and be position vector of AB & BC

\begin{aligned} &(A C)^{2}=(\vec{a})^{2}+(\vec{b})^{2}-0 \\\\ &(A C)^{2}=(\vec{a})^{2}+(\vec{b})^{2} \\\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \end{aligned}

 

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