Question

A,B,C and D are four points such that AB=√24 ; BD=2 ; AD=2√5 ; BC= √150 and AC=√294
1) Show that A, B and C are collinear.
2) Show that A, B, and D form a right angled triangle at D.

Answer 1

Step-by-step explanation:

Let EB=pAB,CE=qCD

then, 0<p,q≤1

Since

EB

+

BC

+

CE

=0

∴pm(2i−6j+2k)+(i−2j)+qn(−6i+15j−3k)=0

⇒(2pm+1−6qn)i+(−6pm−2+15qn)j+(2pm−6qn)k=0

⇒2pm−6qn+1=0,−6pm−2+15qn=0⇒2pm−6qn=0

Solving these, we get

p=

(2m)

1

and q=

(3n).

1

∴0<

(2m)

1

≤ and 0<

(3n)

1

≤1

⇒m≥

2

1

and n≥

3.

1

The area of △BCE=

2

1

∣EB×BC∣=

2m

1

6.