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  • please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 4 sub question 2 maths textbook solution

please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 4 sub question 2 maths textbook solution

Answers (1)

\vec{AB}=4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=4\sqrt{5}

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

Position vector \vec{AB} is given x, y

\vec{AB} = position vector of B - position vector of A

Given:

               A=(-6,3) and B(-2,-5)

Solution:

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 unit vector in

x and y direction

\vec{A}=-6\hat{i}+3\hat{j}\\ \vec{B}=-2i-5\hat{j}

Then the position vector \vec{AB} is given by

\vec{AB} = position vector of B - position vector of A

=\left (2\hat{i}-5\hat{j} \right )-\left (-6\hat{i}+3\hat{j} \right )\\ =-2\hat{i}-5\hat{j} +6\hat{i}-3\hat{j} \\ =4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=\sqrt{4^{2}+(-8)^{2}}=\sqrt{16+64}\\ =\sqrt{80}\\ \left |\vec{AB} \right |=4\sqrt{5}-
 

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