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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 7 Sub Question 2 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products  Exercise 23.1 Question 7 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer: 2\hat{i}-\hat{j}+\hat{k}

Hint: you must know the rules of finding vector from gives dot product values.

Given: \hat{i}-\hat{j}+\hat{k},2\hat{i}+\hat{j}-3\hat{k},    \hat{i}+\hat{j}+\hat{k},     are 4,0 and 2     respectively

Solution:  \hat{i}-\hat{j}+\hat{k} ,2\hat{i}+\hat{j}-3\hat{k},  \hat{i}+\hat{j}+\hat{k}, are 4,0 and 2     respectively

Let a\hat{i}+b\hat{j}+c\hat{k},be the required vector.

Given that

\begin{aligned} &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=4 \\ &\Rightarrow a-b+c=4 \\ &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}-3 \hat{k})=0 \\ \end{aligned}

\begin{aligned} &\Rightarrow 2 a+b-3 c=0 \quad-1 \\ &\Rightarrow 2 a \end{aligned}

\begin{aligned} &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k}) \\\\ &\Rightarrow a+b+c=2 \\ \end{aligned}

From 1 and 2       subtract

\begin{aligned} &(a-b+c)-(a+b+c)=4-2 \\\\ &a-b+c-a-b-c=2 \\\\ &2 b=2 \\\\ &B=-1 \\ \end{aligned}

From 1 and 2       subtract

\begin{aligned} &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ \end{aligned}

\begin{aligned} &\Rightarrow a+b+c=2 \\\\ &(a-b+c)-(a+b+c)=4-2 \\\\ &a-b+c-a-b-c=2 \\ \end{aligned}

2b=2

B=-1

From 1 and 2       subtract

\begin{aligned} \\\\ &(a-b+c)-(2 a+b-3 c)=4-0 \\\\ &a-b+c-2 a-b+3 c=4 \\\\ &-a-2 b+4 c=4 \\\\ &a+2 b-4 c=-4 \\\\ &a=-4-2 b+4 c \end{aligned}

Put b = -1

=-4-2\left ( -1 \right )+4c

=-4+2+4c

A=4c-2

Now from 3

\begin{aligned} &\mathrm{A}+\mathrm{b}+\mathrm{c}=2 \\ &(4 \mathrm{c}-2)+(-1)+\mathrm{c}=2 \\ &4 \mathrm{c}-2-1+\mathrm{c}=2 \\ \end{aligned}

\begin{aligned} &5 \mathrm{c}-3=2 \\ &5 \mathrm{c}=3+2 \\ &5 \mathrm{c}=5 \\ &\mathrm{C}=1 \end{aligned}

Put value of b and c in 1

a-b+c=4

\begin{aligned} &a-(-1)+1=4 \\ &a+1+1=4 \\ &a+2=4 \\ &a=2 \\ &a=2, \quad b=-1, \quad c=1 \\ &\therefore a \hat{\imath}+b \hat{\jmath}+c \hat{k} \cdot 2 \hat{\imath}-\hat{\jmath}+\hat{k} \end{aligned}

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