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Name: Explain solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 8 maths textbook solution
Uploaded: Sun, 2021-08-29 17:44
Description: Explain solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 8 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 8 maths textbook solution

Answers (1)

Answer: $\overrightarrow{d}$ is expressible as linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$

Hint: Express one vector in linear combination of the other two

$\Rightarrow$ Given,

$\begin{aligned} &\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\ &\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k} \\ &\vec{c}=\hat{i}+\hat{j}+\hat{k} \\ &\Rightarrow \vec{a}=x \vec{b}+y \vec{c} \\ &\Rightarrow \hat{i}+2 \hat{j}+3 \hat{k}=x(2 \hat{i}+\hat{j}+3 \hat{k})+y(\hat{i}+\hat{j}+\hat{k}) \\ &\Rightarrow \hat{i}+2 \hat{j}+3 \hat{k}=\hat{i}(2 x+y)+\hat{j}(x+y)+\hat{k}(3 x+y) \end{aligned}$

Comparing coefficients of $\hat{i} ,\hat{j}$ & $\hat{k}$

$2x+y=1$ ...(1)

$x+y=2$ ....(2)

$3x+y=3$ ....(3)

Subtracting (1) and (2)

$2x+y=1$

$\underline{x+y=2}$

$x$ $=-1$

$x+y=2$

$-1+y=2$

$y=3$

Put, x=-1 and y=3

$3x+y=3$

$L.H.S=3x+y$

$=3\left ( -1 \right )+3$

$=-3+3$

$=0$

$\neq R.H.S$

Thus, $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$ are non-coplanar

Now, expressing $\overrightarrow{a}$ as a linear combination of $\overrightarrow{a},\overrightarrow{b}$ & $\overrightarrow{c}$

$\begin{aligned} &\vec{d}=\vec{a} x+\vec{b} y+\vec{c} z \\ &\Rightarrow 2 \hat{i}-\hat{j}-3 \hat{k}=x(\hat{i}+2 \hat{j}+3 \hat{k})+y(2 \hat{i}+\hat{j}+3 \hat{k})+z(\hat{i}+\hat{j}+\hat{k}) \\ &\Rightarrow 2 \hat{i}-\hat{j}-3 \hat{k}=\hat{i}(x+2 y+z)+\hat{j}(2 x+y+z)+\hat{k}(3 x+3 y+z) \end{aligned}$