Question

Show that A (2, 1, 1), B(0, -1, 4), C(4. 3. -2) points are collinear.

Answer 1

Hey there!

Solution;



The Given Points are A ( 2, 1, 1 ), B ( 0, -1, 4 ) and C ( 4, 3, -2 ).


$\vec{A} => 2\vec{i} + \vec{j} + \vec{k}\\ \\ \vec{B} => 0\vec{i} - \vec{j} + 4\vec{k}\\ \\ \vec{C} => 4\vec{i} + 3\vec{j} - 2\vec{k}$


$Now;\\ \\ \vec{AB} => Position\: vector \:of \:B \:- Position\: vector\: of\: A\\ \\ We Get;\\ \\ \vec{AB} => -2\vec{i} + 2\vec{j} + 3\vec{k}\\ \\ \vec{BC} => Position\: vector\: of \:C - Position\: vector\: of \:B$


$We Get;\\ \\ \vec{BC} => 4\vec{i} + 4\vec{j} - 6\vec{k}\\ \\ \vec{CA} => Position\: vector \:of\: A - Position\: vector \:of\: C\\ \\ We Get;\\ \\ \vec{CA} => -2\vec{i} - 2\vec{j} + 3\vec{k}$


$Now; \\ \\ | \vec{AB}|=> \sqrt{(-2)^{2} + (2)^{2} + (3)^{2}}\\ \\ => \sqrt{17} \\ \\ |\vec{BC}| => \sqrt{(4)^{2} + (4)^{2} + (-6)^{2}}\\ \\ => \sqrt{68} \\ \\ |\vec{CA}| => \sqrt{(-2)^{2} + (-2)^{2} + (3)^{2}}\\ \\ => \sqrt{17}\\ \\ Now;\\ \\ |\vec{BC}| = |\vec{AB}| + |\vec{CA}| \\ \\ Substituting \:values;\\ \\ \sqrt{68} = \sqrt{17} + \sqrt{17} \\ \\ 2\sqrt{17} = 2\sqrt{17} \\ \\ Therefore;\\ \\ |\vec{BC}| = |\vec{AB}| + |\vec{CA}|$


Hence, these Points are Collinear.



#TogetherWeGoFar

Answer 2

Answer:

Direction ratio of AB and BC are in proportion,2AB=BC

Thus all points are collinear.

Step-by-step explanation:

To show that points A(2 1, 1), B(0,-1,4) andC (4,3, -2) are collinear.

• Calculate direction ratio of two points taking at a time
• if both direction ratio's are same or in same proportion,than both lines are parallel
• and hence both lines have a point common,i.e.all these points are in one line

now calculate Direction ratio

AB:0-2,-1-1,4-1=-2,-2,3

BC=4-0,3+1,-2-4=4,4,-6

2AB=BC

thus on the basis of direction ratio we can say that both lines are parallel,but in both the lines Point B is common so,we can say that all three points are on one lines.