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Question

Mathematics Question on Vector Algebra

Find the direction cosines of the vector joining the points A(1,2,−3)A(1,2,-3) and B(−1,−2,1)B(-1,-2,1) directed from AA to BB.

Answer

The correct answer is:(−13,−23,23).(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}).
The given points are A(1,2,−3)A(1,2,-3) and B(−1,−2,1).B(-1,-2,1).
∴AB⃗=(−1−1)i^+(−2−2)j^+1−(−3)k^∴\vec{AB}=(-1-1)\hat{i}+(-2-2)\hat{j}+{1-(-3)}\hat{k}
⇒AB⃗=−2i^−4j^+4k^⇒\vec{AB}=-2\hat{i}-4\hat{j}+4\hat{k}
∴∣AB⃗∣=(−2)2+(−4)2+42=4+16+16=36=6∴|\vec{AB}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6
Hence,the direction cosines of AB⃗\vec{AB} are (−26,−46,46)=(−13,−23,23).(\frac{-2}{6},\frac{-4}{6},\frac{4}{6})=(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}).