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  • Provide Solution for RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 13

Provide Solution for RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 13

Answers (1)

best_answer

Answer:

-2

Hint:

Use the formula  \overrightarrow{A B}=K \overrightarrow{B C}  to prove it is collinear

Given:

\begin{aligned} &A=\lambda \hat{i}-10 \hat{j}+3 \hat{k} \\ &B=\hat{i}-\hat{j}+3 \hat{k} \\ &C=3 \hat{i}+5 \hat{j}+3 \hat{k} \end{aligned}

Solution:

\begin{aligned} &A=\lambda \hat{i}-10 \hat{j}+3 \hat{k} \\ &B=\hat{i}-\hat{j}+3 \hat{k} \\ &C=3 \hat{i}+5 \hat{j}+3 \hat{k} \end{aligned}
\begin{aligned} &\overrightarrow{A B}=(\hat{i}-\hat{j}+3 \hat{k})-(\lambda \hat{i}-10 \hat{j}+3 \hat{k}) \\ &=(1-\lambda) \hat{i}+(-1+10) \hat{j}+(3-3) \hat{k} \\ &=(1-\lambda) \hat{i}+9 \hat{j} \end{aligned}
\begin{aligned} &\overrightarrow{B C}=(3 \hat{i}+5 \hat{j}+3 \hat{k})-(\hat{i}-\hat{j}+3 \hat{k}) \\ &=(3-1) \hat{i}+(5+1) \hat{j}+(3-3) \hat{k} \\ &=2 \hat{i}+6 \hat{j} \\ &\overline{A B}=K \overline{B C} \\ &(1-\lambda) \hat{i}+9 \hat{j}=K(2 \hat{i}+6 \hat{j}) \end{aligned}

Comparing

\begin{aligned} &6 K=9 \\ &\therefore K=\frac{9}{6}=\frac{3}{2} \\ &K=\frac{3}{2} \\ &1-\lambda=2 K \end{aligned}
\begin{aligned} &1-\lambda=2\left(\frac{3}{2}\right) \\ &1-\lambda=3 \\ &1-3=\lambda \\ &\lambda=-2 \end{aligned}

 

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