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Provide solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Fill in the blanks question 10

Answers (1)

Answer:- x+y+z=2

Hint:-To solve this equation, we use substitute method

Given:- If the vector \vec{a}=2 \hat{i}-(y+z) \hat{j}+5 \hat{k} \text { and } \vec{b}=(x+y) \hat{i}+3 \hat{j}+(z+x) \hat{k} are then x+y+z=____

Solution:- \vec{a}=\vec{b}

2 \hat{i}-(y+z) \hat{j}+5 \hat{k}=(x+y) \hat{i}+3 \hat{j}+(z+x) \hat{k}

Comparing we get

\begin{array}{ll} x+y=2 & \text {...( } i) \\\\ -(y+z)=3 & \text {...(ii) } \\\\ , z+x=5 & \text {...(iii) } \end{array}

From eq (ii) we make y the subject to the equation

Substitute for y=-3-z  in the equation(i)

\begin{aligned} &x-3-z=2 \quad ......(iv)\\ \\ &x-z=5 \end{aligned}

Solving eq(iii) & (iv)

\begin{aligned} &z+x+x-z=5+5 \\\\ &2 x=10 \\\\ &x=5 \end{aligned}

Put the values of x in eq(iii)

\begin{aligned} &z+5=5 \\\\ &z=0 \end{aligned}

 Put the values of z in eq(ii)

\begin{aligned} &\mathrm{y}=-3 \\\\ &\text { so, } \mathrm{x}+\mathrm{y}+\mathrm{z}=5+(-3)+0 \\\\ &\quad=5-3 \\\\ &\quad=2 \end{aligned}

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