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Provide solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise Fill in the blanks question 10

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Final Answer6 i^{\wedge}-9j^{\wedge}+18 k^{\wedge}

Hint:

Use direction ratio to find direction cosine.

Given:

|\vec{r}|=21 \text { and }(a, b, c)=(2,-3,6)

To Find: 

vector \vec{r}

Solution:

Direction cosine is related to Direction ratio as

l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}

l=\frac{2}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}, m=\frac{-3}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}, n=\frac{6}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}

\begin{aligned} &l=\frac{2}{7}, m=\frac{-3}{7}, n=\frac{6}{7} \\\\ \end{aligned}

\vec{r}=|\vec{r}|\left(l i^{\wedge}+m_{j}^{\wedge}+n k^{\wedge}\right)

\begin{aligned} &\text { Now } \vec{r}=21\left(\frac{2}{7} i^{\wedge}+\frac{(-3)}{7} \hat{\jmath}+\frac{6 k^{\wedge}}{7}\right) \\\\ &\vec{r}=3\left(2 i^{\wedge}-3 \hat{\jmath}+6 k^{\wedge}\right) \\\\ &\vec{r}=6 i^{\wedge}-9 \hat{\jmath}+18 k^{\wedge} \end{aligned}

Therefore, the vector \vec{r} \text { is }\left(6 i^{\wedge}-9 j^{\wedge}+18 k^{\wedge}\right)

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