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Provide Solution for RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 8

Answers (1)

Answer:

$1$

Hint:

Represent point in form of vectors and use $\overrightarrow{A B}=\lambda \overrightarrow{B C}$

Given:

Points $A(m,-1), B(2,1), C(4,5)$ are collinear

Solution:

$\begin{aligned} &A=m \hat{i}-\hat{j} \\ & \end{aligned}$

$B=2 \hat{i}+\hat{j} \\$

$C=4 \hat{i}+5 \hat{j} \\$

$\overrightarrow{A B}=(2 \hat{i}+\hat{j})-(m \hat{i}-\hat{j}) \\$

$=(2-m) \hat{i}+(1+1) \hat{j} \\$

$=(2-m) \hat{i}+2 \hat{j} \\$

$\overrightarrow{B C}=(4 \hat{i}+5 \hat{j})-(2 \hat{i}+\hat{j}) \\$

$=(4-2) \hat{i}+(5-1) \hat{j} \\$

$=2 \hat{i}+4 \hat{j}$

If they are collinear then

$\begin{aligned} &\overrightarrow{A B}=\lambda \overrightarrow{B C} \\ &(2-m) \hat{i}+2 \hat{j}=\lambda(2 \hat{i}+4 \hat{j}) \end{aligned}$

Comparing

$\begin{aligned} &2=\lambda(4) \\ &\lambda=\frac{1}{2} \\ &(2-m)=\lambda(2) \end{aligned}$

Put the value of $\lambda$

$\begin{aligned} &2-m=\frac{1}{2} \times 2 \\ &2-m=1 \\ &m=2-1 \\ &m=1 \end{aligned}$

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