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  • Explain Solution R.D. Sharma Class 12 Chapter 22 Algebra of Vectors Exercise Very Short Answer Type Question 36 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 22 Algebra of Vectors  Exercise Very Short Answer Type Question 36 Maths Textbook Solution.

Answers (1)

Answer: \text { Thus, } \vec{a} \text { is parallel to } \vec{b} \text { and hence in the same direction. }

Hint: You must know the rules of vector functions

Given: Write two different factors having same direction

Solution:

           Let \vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}, \vec{b}=4 \hat{i}+2 \hat{j}+4 \hat{k}

The direction cosines of \hat{a} is

\begin{aligned} &l=\frac{2}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{2}{\sqrt{9}}=\frac{2}{3} \\ &m=\frac{1}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{1}{\sqrt{9}}=\frac{1}{3} \\ &n=\frac{2}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{2}{\sqrt{9}}=\frac{2}{3} \\ \end{aligned}

And direction cosines of \vec{b} is

\begin{aligned} &l=\frac{4}{\sqrt{(4)^{2}+(2)^{2}+(4)^{2}}}=\frac{4}{\sqrt{16+4+16}}=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3} \\ &m=\frac{2}{\sqrt{(4)^{2}+(2)^{2}+(4)^{2}}}=\frac{2}{\sqrt{16+4+16}}=\frac{2}{\sqrt{36}}=\frac{2}{6}=\frac{1}{3} \\ &n=\frac{4}{\sqrt{(4)^{2}+(2)^{2}+(4)^{2}}}=\frac{4}{\sqrt{16+4+16}}=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3} \end{aligned}

The direction cosines of \vec{a} and \vec{b} are same.

\text { Thus, } \vec{a} \text { is parallel to } \vec{b} \text { and hence in the same direction. }

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