Question

find the acute angle between the lines AB and CD where A=(2,3,-5),B=(10,7,-4),C=(0,0,2),D=(3,3,-4)​

Answer 1

$\theta =cos^{-1}(\frac{10}{9\sqrt{6} })$

Step-by-step explanation:

Given , A (2,3,-5) ,B(10,7,-4), C(0,0,2) and(3,3,-4)

The direction ratio of line AB is (10-2,7-3,-4+5)=(8,4,1)

The direction ratio of line CD is(3-0,3-0,-4-2)=(3,3,-6)

Let $\theta$ be the angle between the lines .

Therefore,   $cos \theta$

           $=\frac{8.3+4.3+1.(-6)}{\sqrt{8^2+4^2+1^2}\sqrt{3^2+3^2+(-6)^2} }$

            $=\frac{30}{\sqrt{81} \sqrt{54} }$

           $=\frac{10}{9\sqrt{6} }$

Therefore       $\theta =cos^{-1}(\frac{10}{9\sqrt{6} })$      

Answer 2

$\theta=cos^{-1}(\frac{5}{9\sqrt 2})^{\circ}$

Step-by-step explanation:

A=(2,3,-5)

B=(10,7,-4)

C=(0,0,2)

D=(3,3,-4)

AB=B-A=(10,7,-4)-(2,3-5)=(8,4,1)

CD=D-C=(3,3-4)-(0,0,2)=(3,3,-6)

$\mid AB\mid=\sqrt{(8^2+4^2+1^2}=9 units$

Using the formula

$\mid r\mid=\sqrt[x^2+y^2+z^2}$$\sqrt[x^2+y^2+z^2}$$\sqrt{x^2+y^2+z^2}$

Where vector r=xi+yj+zk

$\mid CD\mid=\sqrt{3^2+3^2+(-6)^2}=6\sqrt 2 units$

$AB\cdot CD=<8,4,1>\cdot <3,3-6>=24+12-6=30$

$cos\theta=\frac{AB\cdot CD}{\mid AB\mid\mid CD\mid}$

Using the formula

$cos\theta=\frac{30}{9\times 6\sqrt 2}=\frac{5}{9\sqrt 2}$

$\theta=cos^{-1}(\frac{5}{9\sqrt 2})^{\circ}$

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