Show that the four points A, B, C, D with position vectors a, vector b, vector c, vector d respectively such that 3vector a- 2 vector b+

Question

Show that the four points A, B, C, D with position \(\vec a,\vec b, \vec c, \vec d\) respectively such that 3\(\vec a\)- 2 \(\vec b\)+ 5\(\vec c\)- 6\(\vec d\)= \(\vec 0\) are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

Answer 1

Given the position vectors of points A, B, C and D are \(\vec a, \vec b, \vec c\)and \(\vec d\) respectively.

We have \(3\vec a-2\vec b+5\vec c-6\vec d=0\)

Rearranging the terms in the above equation,

\(3\vec a+5\vec c=\vec 2b+6\vec d\)

Observe that the sum of coefficients on the LHS of this equation (3 + 5 = 8) is equal to that on the RHS (2 + 6 = 8).

We now divide the equation with 8 on both sides.

Now, consider the LHS of this equation.

Let 

the position vector of some point X.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors \(\vec a\)and \(\vec b\) respectively, internally in the ratio m : n is

Here, m = 3 and n = 5

So, X divides CA internally in the ratio 3 : 5.

Similarly, considering the RHS of this equation, we have the same point X dividing DB in the ratio 2 : 6.

So, the point X lies on both the line segments AC and BD making it the point of intersection of AC and BD.

As AC and BD are two straight lines having a common point, we have all the points A, B, C and D lying in the same plane.

Thus, the points A, B, C and D are coplanar and in addition, the position vector of the point of intersection of line segments AC and BD is