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

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

Answers (1)

Answer:

AB and CD intersect at the point P

Hint:

If \overrightarrow{A B} \& \overrightarrow{C D}   intersect at P, it makes A-P-B collinear and C-P-D collinear. So prove they are collinear  

Given:

\begin{aligned} &A=-2 \hat{i}+3 \hat{j}+5 \hat{k} \\ & \end{aligned}
B=7 \hat{i}+0 \hat{j}-\hat{k}=7 \hat{i}-\hat{k} \\
C=-3 \hat{i}-2 \hat{j}-5 \hat{k} \\
D=3 \hat{i}+4 \hat{j}+7 \hat{k} \\
P=1 \hat{i}+2 \hat{j}+3 \hat{k}

Solution:

\begin{aligned} &A=-2 \hat{i}+3 \hat{j}+5 \hat{k} \\ &B=7 \hat{i}+0 \hat{j}-\hat{k}=7 \hat{i}-\hat{k} \\ &C=-3 \hat{i}-2 \hat{j}-5 \hat{k} \\ &D=3 \hat{i}+4 \hat{j}+7 \hat{k} \\ &P=1 \hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}

\overrightarrow{A P}=  Position vector of P – Position vector of A

\begin{aligned} &=(\hat{i}+2 \hat{j}+3 \hat{k})-(-2 \hat{i}+3 \hat{j}+5 \hat{k})\\ &=(1+2) \hat{i}+(2-3) \hat{j}+(3-5) \hat{k}\\ &=3 \hat{i}-\hat{j}-2 \hat{k} \end{aligned}                           … (i)

\overrightarrow{P B}  Position vector of B – Position vector of P

\begin{aligned} &=(7 \hat{i}-\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\\ & \end{aligned}

=(7-1) \hat{i}+(0-2) \hat{j}+(-1-3) \hat{k}\\

=6 \hat{i}-2 \hat{j}-4 \hat{k}\\

=2(3 \hat{i}-\hat{j}-2 \hat{k}) \\         cgfy… (ii)

\overline{P B}=2 \overrightarrow{A P}

Thus, A-P-B are collinear

\overrightarrow{C P}=  Position vector of P – Position vector C

\begin{aligned} &=(\hat{i}+2 \hat{j}+3 \hat{k})-(-3 \hat{i}-2 \hat{j}-5 \hat{k}) \\ &=(1+3) \hat{i}+(2+2) \hat{j}+(3+5) \hat{k} \\ &=4 \hat{i}+4 \hat{j}+8 \hat{k} \\ &=4(\hat{i}+\hat{j}+2 \hat{k}) \end{aligned}

\overrightarrow{P D}=  Position vector of D – Position vector of P

\begin{aligned} &=(3 \hat{i}+4 \hat{j}+7 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\ &=(3-1) \hat{i}+(4-2) \hat{j}+(7-3) \hat{k} \\ &=2 \hat{i}+2 \hat{j}+4 \hat{k} \\ &=2(\hat{i}+\hat{j}+2 \hat{k}) \\ &\overrightarrow{P D}=\frac{2}{4} \overrightarrow{C P} \\ &\overrightarrow{P D}=\frac{1}{2} \overrightarrow{C P} \end{aligned}

Thus, \overrightarrow{P D} \& \overline{C P}  are collinear

If A-P-B are collinear and also C-P-D are collinear

Hence  \overrightarrow{A B} \text { and } \overrightarrow{C D}  intersect at P

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