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  • Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 1 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 1 Maths Textbook Solution.

Answers (1)

Answer:

Point A, B and C are collinear

Hint:

Prove, that the position vectors are parallel to each other, so that they have one common point

Given:

Point A, B and C with position vectors  \vec{a}-2 \vec{b}+3 \vec{c}, 2 \vec{a}+3 \vec{b}-4 \vec{c} \: \text { and }-7 \vec{b}+10 \vec{c}   respectively

Solution:

Let O be the point of origin for the position vectors

\begin{aligned} &\therefore \overrightarrow{O A}=(\vec{a}-2 \vec{b}+3 \vec{c})-(0+0+0)=\vec{a}-2 \vec{b}+3 \vec{c} \\\\ &\text { Similarly } \overrightarrow{O B}=2 \vec{a}+3 \vec{b}-4 \vec{c} \text { and } \overrightarrow{O C}=-7 \vec{b}+10 \vec{c} \end{aligned}

Now, \overrightarrow{A B}=  Position vector B – position vector A

\begin{aligned} &=\overline{O B}-\overline{O A}\\ & \end{aligned}

=(2 \vec{a}+3 \vec{b}-4 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c})\\

=[(2-1) \vec{a}]+[(3+2) \vec{b}]+[(-4-3) \vec{c}]\\

\overrightarrow{A B}=\vec{a}+5 \vec{b}-7 \vec{c}                           …(i)

\overrightarrow{BC}  Position of vector C – Position of vector B

\begin{aligned} &\overrightarrow{B C}=[-7 \vec{b}+10 \vec{c}]-[2 \vec{a}+3 \vec{b}-4 \vec{c}] \\ & \end{aligned}

=[(0-2) \vec{a}]+[(-7-3) \vec{b}]+[(10+4) \vec{c}] \\

=-2 \vec{a}-10 \vec{b}+14 \vec{c}

=-2(\vec{a}+5 \vec{b}-7 \vec{c})                         …(ii)

Substituting (i) in (ii)

\overrightarrow{B C}=-2(\overline{A B})

Thus, \overrightarrow{AB }\: and \: \overrightarrow{B C}  are parallel vectors

As both vectors have one common point in B

\overrightarrow{AB }\: and \: \overrightarrow{B C}  are collinear

Thus, point A, B and C are collinear

 

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