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  • please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 1 maths textbook solution

please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 1 maths textbook solution

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5

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

Given: Position vector of a point (-4,-3) 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}=(-4)\hat{i} +(-3)\hat{j}

               Now we need to find magnitude of \vec{a} i.e \left |\vec{a} \right |

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

               Here, x=-4 and y=-3

\left |\vec{a} \right |=\sqrt{(-4)^{2} +(-3)^{2}}\\ \left |\vec{a} \right |=\sqrt{16+9}\\ \left |\vec{a} \right |=\sqrt{25}\\ \left |\vec{a} \right |=5

thus \left |\vec{a} \right |=5

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