Question:
Find the direction cosines of the vector joining the points \(A(1,2,-3)\) and \(B(-1,-2,1)\) directed from \(A\) to \(B\).
Find the direction cosines of the vector joining the points \(A(1,2,-3)\) and \(B(-1,-2,1)\) directed from \(A\) to \(B\).
Updated On: Sep 19, 2023
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Solution and Explanation
The correct answer is:\((\frac{-1}{3},\frac{-2}{3},\frac{2}{3}).\)
The given points are \(A(1,2,-3)\) and \(B(-1,-2,1).\)
\(∴\vec{AB}=(-1-1)\hat{i}+(-2-2)\hat{j}+{1-(-3)}\hat{k}\)
\(⇒\vec{AB}=-2\hat{i}-4\hat{j}+4\hat{k}\)
\(∴|\vec{AB}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6\)
Hence,the direction cosines of \(\vec{AB}\) are \((\frac{-2}{6},\frac{-4}{6},\frac{4}{6})=(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}).\)
The given points are \(A(1,2,-3)\) and \(B(-1,-2,1).\)
\(∴\vec{AB}=(-1-1)\hat{i}+(-2-2)\hat{j}+{1-(-3)}\hat{k}\)
\(⇒\vec{AB}=-2\hat{i}-4\hat{j}+4\hat{k}\)
\(∴|\vec{AB}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6\)
Hence,the direction cosines of \(\vec{AB}\) are \((\frac{-2}{6},\frac{-4}{6},\frac{4}{6})=(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}).\)
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