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please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 2 maths textbook solution

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n=\pm 5

Hints: \left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |

Given: Position vector of \vec{a} point (12,n) be \left | \vec{a} \right |

               Find: \left | \vec{a} \right |

We know position vector of a point(x,y) is given by x\hat{i} +y\hat{j} where \hat{i} and \hat{j} are units vectors in  x and y directions.

               \vec{a}=12\hat{i} +n\hat{j}

Then:

               \left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}

               Here, x=12 and y=n

\left |\vec{a} \right |=\sqrt{12^{2} +n^{2}}\\ \left |\vec{a} \right |=13[given]

13=\sqrt{(12)^{2} +(n)^{2}}\\

Squaring both sides

13^{2}=(12)^{2} +(n)^{2}\\ 169=144+n^{2}\\ n^{2}=25\\ n=\pm 5

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