0 votes
1.5k views
in Vectors by (29.5k points)
closed by

Prove that the points having position vectors \(\hat i+2\hat j+3\hat k\)\(3\hat i+4\hat j+7\hat k\) \(-3\hat i-2\hat j-5\hat k\) are collinear.

1 Answer

+1 vote
by (30.5k points)
selected by
 
Best answer

Let us understand that, two more points are said to be collinear if they all lie on a single straight line

 Let the points be A, B and C having position vectors such that,

Position vector of A = \(\vec i+2\vec j+3\vec k\)

Position vector of B = \(3\vec i+4\vec j+7\vec k\)

 Position vector of C = \(-3\vec i-2\vec j-7\vec k\)

So, in this case if we prove that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore\(\vec{AB}\) is given by

\(\vec{AB}\) = Position vector of C - Position vector of B

Let us note the relation between \(\vec{AB}\) and \(\vec{BC}\).

We know,

This relation shows that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other.

But also, \(\vec B\) is the common vector in \(\vec{AB}\) and \(\vec{BC}\).

⇒ \(\vec{AB}\) and \(\vec{BC}\) are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...