Question #128658

3x-3=3y-6 =z-7,find the unit vector parallel to the above line

1
Expert's answer
2020-08-06T17:42:50-0400

Let us rewrite the equation in form

x113=y213=z71\dfrac{x-1}{\frac{1}{3}} = \dfrac{y-2}{\frac{1}{3}} = \dfrac{z-7}{1} .

In such form the denominators represent the coordinates of the vector parallel to the line. The coordinates are (1/3, 1/3, 1). But we can see the vector is not unit, so we should normalize it.

The length of the vector above is

(1/3)2+(1/3)2+12=113.\sqrt{(1/3)^2+(1/3)^2+1^2} = \dfrac{\sqrt{11}}{3}.

Therefore, the unit vector will have coordinates (1/311/3,1/311/3,111/3)\left(\dfrac{1/3}{\sqrt{11}/3}\,,\dfrac{1/3}{\sqrt{11}/3}\,, \dfrac{1}{\sqrt{11}/3}\right) , after simplifying

(1111,1111,31111)\left( \dfrac{\sqrt{11}}{11}\,, \dfrac{\sqrt{11}}{11}\,, \dfrac{3\sqrt{11}}{11}\right) .


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