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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 1 Maths Textbook Solution.

Answers (1)

Answer:  \hat{i}+2\hat{j}+\hat{k}

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

Given: dot product of \hat{i}+\hat{j}-3\hat{k},\hat{i}+3\hat{j}-2\hat{k}      and  2\hat{i}+\hat{j}+4\hat{k} are 0,5 and 8 respectively,

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

Give that,

\begin{aligned} &\left(a \hat{\imath}+\widehat{b}_{j}+c \hat{k}\right) \cdot(\hat{\imath}+\hat{\jmath}-3 \hat{k})=0 \\ &a+b-3 c=0 \quad-----(1) \\ &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}-3 \hat{k})=5 \\ &a+3 b-2 c=5 \quad-----(2) \\ &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}+4 \hat{k})=8 \\ &2 a+b+4 c=8 \quad-----(3) \end{aligned}

From 1 and 2, subtract

                                                \begin{aligned} &(a+b-3 c)-(a+3 b-2 c)=0-5\\ \end{aligned}

\begin{aligned} &a+b-3 c-a-3 b-2 c=-5\\ &-2 b-c=-5\\ &c=5-2 b\\ \end{aligned}

From 2 and 3, subtract

                                        \begin{aligned} &(a+3 b-2 c)-(2 a+3 b+4 c)=5-8\\ \end{aligned}

\begin{aligned} &a+3 b-2 c-2 a-b-4 c=-3\\ &-a+2 b-6 c=-3\\ &a-2 b+6 c=3\\ &a=3+2 b-6 c \end{aligned}

Where c=5-2b

\begin{aligned} &a=3+2 b-6 c(5-2 b) \\ &=3+2 b-3 c+12 b \\ &a=27+14 b \end{aligned}

Put value of (a) and (c) in 1

\begin{aligned} &A+b-3 c=0 \\ \end{aligned}

\begin{aligned} &(-27+14 b)+b-3(5-2 b)=0 \\\\ &-27+14 b+b-15+6 b=0 \\ \end{aligned}

\begin{aligned} &21 b-42=0 \end{aligned}

21b=42

B=\frac{42}{21}

B=2

Now put value of b in (c) and (a)

c=5-2b

c=5-2\times 2

=5-4

c=1

and a=-27+14b

=-27+14\left ( 2 \right )

=-27+28

=-27+28

a=1

a=1,b=2,c=1

So,a \hat{\imath}+b \hat{\jmath}+c \hat{k}=\hat{\imath}+2 \hat{\jmath}+\hat{k}

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