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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 37 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products Exercise 23.1 Question 37 Maths Textbook Solution.

Answers (1)

Answer:  -\hat{i}-\hat{j}-\hat{k},7\hat{i}-2\hat{j}-5\hat{k}

Hint: You must know the rules of solving vectors.

Given: Decompose the vector 6\hat{i}-3\hat{j}-6\hat{k} into vector which are parallel and perpendicular to the vector \hat{i}+\hat{j}+\hat{k}

Solution:

Let \vec{a}=6\hat{i}-3\hat{j}-6\hat{k}

\vec{b}=\hat{i}+\hat{j}+\hat{k}

And \vec{x} and \vec{y}  be such that,

                                        \Rightarrow \vec{a}=\vec{x}+\vec{y}

\Rightarrow \vec{y}=\vec{a}-\vec{x}

Since  \vec{x} is parallel to \vec{b}

\begin{aligned} &\Rightarrow \vec{x}=t \vec{b} \\ &\Rightarrow \vec{x}=t(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ &\Rightarrow \vec{x}=t \hat{\imath}+t \hat{\jmath}+t \hat{k} \end{aligned}

Substituting the values of \vec{x} and  \vec{a}

\begin{aligned} &\vec{y}=(6 \hat{i}-3 \hat{\jmath}-6 \hat{k})-(t \hat{\imath}+t \hat{\jmath}+t \hat{k}) \\ &\vec{y}=(6-t) \hat{\imath}+(-3-t) \hat{\jmath}+(-6-t) \hat{k} \end{aligned}

Since y is perpendicular to \vec{b}

\begin{aligned} &\Rightarrow \vec{y} \cdot \vec{b}=0 \\ &\Rightarrow[(6-t) \hat{\imath}+(-3-t) \hat{\jmath}+(-6-t) \hat{k}] \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=0 \\ &\Rightarrow 1(6-t)+1(-3-t)+1(-6-t)=0 \\ \end{aligned}

\begin{aligned} &\Rightarrow-3-3 t=0 \\ &\Rightarrow t=-1 \\ &\vec{x}=-\hat{\imath}-\hat{\jmath}-\hat{k} \\ &\vec{y}=7 \hat{\imath}-2 \hat{\jmath}-5 \hat{k} \end{aligned}

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