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  • Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 11 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.7 Question 11 Maths Textbook Solution.

Answers (1)

best_answer

Answer:

2:3

Hint:
When, in a three collinear points A, B and C, if B divides AC then formula is  \frac{n \vec{a}+m \vec{b}}{n+m}

Given:

A(1,-2,-8), B(5,0,-2) \& C(11,3,7)  Are collinear points

Solution:

\begin{aligned} &A=\hat{i}-2 \hat{j}-8 \hat{k} \\ &B=5 \hat{i}-2 \hat{k} \\ &C=11 \hat{i}+3 \hat{j}+7 \hat{k} \\ &\overrightarrow{A B}=(5 \hat{i}-2 \hat{k})-(\hat{i}-2 \hat{j}-8 \hat{k}) \\ &=(5-1) \hat{i}+(0+2) \hat{j}+(-2+8) \hat{k} \end{aligned}

=4 \hat{i}+2 \hat{j}+6 \hat{k} \\                                            … (i)

\overrightarrow{B C}=(11 \hat{i}+3 \hat{j}+7 \hat{k})-(5 \hat{i}-2 \hat{k}) \\

=(11-5) \hat{i}+(3-0) \hat{j}+(7+2) \hat{k} \\

\begin{aligned} & &=6 \hat{i}+3 \hat{j}+9 \hat{k} \end{aligned}                                         … (ii)

Rewriting (i) and (ii)

\overrightarrow{A B}=2(2 \hat{i}+\hat{j}+3 \hat{k}) \\                          … (iii)

\begin{aligned} & &\overrightarrow{B C}=3(2 \hat{i}+\hat{j}+3 \hat{k}) \end{aligned}                         … (iv)

From (iii)

\overrightarrow{B C}=\frac{3}{2} \overline{A B}

Thus, \overrightarrow{B C} \& \overrightarrow{A B}  are parallel to each other

Therefore they are collinear

Now,

 

Now, if B is dividing AC internally into m and n then,

\begin{aligned} &B=\left(\frac{11 m+n}{m+n}\right) \hat{i}+\left(\frac{3 m-2 n}{m+n}\right) \hat{j}+\left(\frac{7 m-8 n}{m+n}\right) \hat{k} \\\\ &5 \hat{i}-2 \hat{k}=\left(\frac{11 m+n}{m+n}\right) \hat{i}+\left(\frac{3 m-2 n}{m+n}\right) \hat{j}+\left(\frac{7 m-8 n}{m+n}\right) \hat{k} \end{aligned}

Comparing,

\begin{aligned} &5=\frac{11 m+n}{m+n} \\ &5 m+5 n=11 m+n \\ &\therefore 6 m-4 n=0 \end{aligned}                         … (v)

\begin{gathered} 0=\frac{3 m-2 n}{m+n} \\ 0=3 m-2 n \\ 3 m-2 n=0 \end{gathered}                            … (vi)

\begin{aligned} &-2=\frac{7 m-8 n}{m+n} \\ &-2 m-2 n=7 m-8 n \\ &9 m-6 n=0 \end{aligned}                            … (vii)

Now, solving equation (v) and (VI) by elimination method

\begin{aligned} &6 m-4 n=0 \\ &2(3 m-2 n)=0 \\ &3 m-2 n=0 \\ &3 m=2 n \end{aligned}

If we suppose n=1 then

3 m=2(1) \\

\begin{aligned} & &m=\frac{2}{3} \end{aligned}  For all the three equation

Posted by

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