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Need solution for RD Sharma Maths Class 12 Chapter 22 Algera of Vectors Excercise 22.7 Question 4

Answers (1)

Answer:

$8$

Hint:

As the given vectors are collinear, then use formula $\overrightarrow{A B}=\lambda \overrightarrow{B C}$

Given:

Three vectors are collinear.

Solution:

Let,

$\begin{aligned} &A=10 \hat{i}+3 \hat{j} \\ \end{aligned}$

$B=12 \hat{i}-5 \hat{j} \\$

$C=a \hat{i}+11 \hat{j}$

So the above three points A, B and C represents three vectors collinear to each other

$\begin{aligned} &\overrightarrow{A B}=(12 \hat{i}-5 \hat{j})-(10 \hat{i}+3 \hat{j}) \\ & \end{aligned}$

$=(12-10) \hat{i}+(-5-3) \hat{j} \\$

$=2 \hat{i}-8 \hat{j} \\$

$\overrightarrow{B C}=(a \hat{i}+11 \hat{j})-(12 \hat{i}-5 \hat{j}) \\$

$=(a-12) \hat{i}+(11+5) \hat{j} \\$

$=(a-12) \hat{i}+16 \hat{j}$

As, A,B and C are collinear, then there exists $\overrightarrow{A B}=\lambda \overrightarrow{B C}$

$\therefore(2 \hat{i}-8 \hat{j})=\lambda(a-12) \hat{i}+\lambda 16 \hat{j}$

Comparing

$\begin{aligned} &-8 j=\lambda 16 \hat{j} \\ & \end{aligned}$

$-8=\lambda 16 \\$

$\lambda=\frac{-1}{2} \\$

$2 \hat{i}=\lambda(a-12) \hat{i} \\$

$2=\lambda(a-12)$

Put the value of $\lambda$

$\begin{aligned} &2=\frac{-1}{2}(a-12) \\ \end{aligned}$

$-4=a-12 \\$

$a=8$

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