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  • Please solve RD Sharma class 12 chapter 26 Direction Cosines and Direction Ratios exercise Fill in the blanks question 13 maths textbook solution

Please solve RD Sharma class 12 chapter 26 Direction Cosines and Direction Ratios exercise Fill in the blanks question 13 maths textbook solution

Answers (1)

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Final Answer:  Point P=(-2, 4,-4)
Hint:

 Use direction ratio to find direction cosine

Given:
\overrightarrow{\mid O P} \mid=6\; and\\\\ (a, b, c)=(-1,2,-2)

To Find: 

Point P

Solution:

Direction cosine is related to direction ratio as :

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

l=\frac{-1}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, m=\frac{2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, n=\frac{-2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}

l=\frac{-1}{3}, m=\frac{2}{3}, n=\frac{-2}{3}

Now, \overrightarrow{O P}= Position vector of P - Position vector of O

\Rightarrow \overrightarrow{\mathrm{1}}=x i^{\wedge}+y \mathrm{l}^{\wedge}+z k^{\wedge}

where x,y,z are the coordinate of P
Now \overrightarrow{O P}=|\overrightarrow{O P}|\left(l \hat{\imath}+m \hat{j}+n k^{n}\right)

\begin{aligned} &\left(x i^{\wedge}+y i^{\wedge}+2 k^{\wedge}\right)=6\left(-\frac{1}{3} i^{\wedge}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k^{\wedge}\right) \\\\ &\left.\left(x i^{\wedge}+y\right)^{\wedge}+2 k^{\wedge}\right)=\left(-2 \hat{1}+4 \hat{\jmath}-4 k^{\wedge}\right) \end{aligned}

Comparing both

(x, y, z)=(-2,4,-4)

Therefore, the coordinate of point P are (-2,4,-4).

 

Posted by

infoexpert26

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