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  • Need solution for RD sharma maths class 12 chapter 22 Algebra of Vector exercise 22.5 question 11

Need solution for RD sharma maths class 12 chapter 22 Algebra of Vector exercise 22.5 question 11

Answers (1)

\lambda=8 and \mu =-5

Hint: here we use sector formula

 \frac{\left ( mx_{2}+nx_{1} \right )}{m+n},\frac{\left ( my_{2}+ny_{1} \right )}{m+n}

Where m,n are the given ratios

Given: the position vectors of points A, B and C are

\vec{A}=\lambda \hat{i}+3\hat{j}\\\vec{B}=12\hat{i}+\mu \hat{j}\\\vec{C}=11\hat{i}-3\hat{j}

Solution:

It is given that C, divides the line segment joining A and B in the ratio 3:1

\begin{aligned} &11 \hat{\imath}-3 \hat{\jmath}=3 \times(12 \hat{\imath}+\mu \hat{\jmath})+1(\lambda \hat{\imath}+3 \hat{\jmath}) \\ &11 \hat{\imath}-3 \hat{\jmath}=\frac{36 \hat{\imath}+3 u \hat{\jmath}+\lambda \hat{\imath}+3 \hat{\jmath}}{4} \\ \end{aligned}

\begin{aligned}&11 \hat{\imath}-3 \hat{\jmath}=\frac{(36+\lambda) \hat{\imath}+(3 \mu+3) \hat{\jmath}}{4} \\ &4(11 \hat{\imath}-3 \hat{\jmath})=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \\ &44 \hat{\imath}-12 \hat{\jmath}=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \end{aligned}
Equating the corresponding components, we get;

36+\lambda=44\\ \lambda=44-36\\ \lambda=8
and
3\mu+3=-12\\ 3\mu=-12-3=-15\\ \mu=\frac{-15}{3}=-5\\ \mu=-5

Thus, the values of \lambda and \mu are 8 and -5 respectively            

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