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  • please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 4 sub question 1 maths textbook solution

please solve RD sharma class 12 chapter 22 Algebra of vector exercise 22.5 question 4 sub question 1 maths textbook solution

Answers (1)

\vec{AB}=-3\hat{i}+4\hat{j}\\ \left |\vec{AB} \right |=5

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

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

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

Given:

               A=(4,-1) and B(1,3)

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

A=4\hat{i}+(-1)\hat{j}\\ B=i+3\hat{j}

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

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

=\left (\hat{i}+3\hat{j} \right )-\left (4\hat{i}-\hat{j} \right )\\ =\hat{i}+3\hat{j} -4\hat{i}+\hat{j} \\ AB=-3\hat{i}+4j
So,
\left |\vec{AB} \right |=(-3)^{2}+(4)^{2}\\ =\sqrt{9+16}=\sqrt{25}\\\left |\vec{AB} \right |=5

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