The vertices A B C of a triangle are (2,-1,-3) (4,2,3) and (6,3,4) respectively. Show that vector AB=(2,3,6) and AC=9.
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Step-by-step explanation:
The vertices of ΔABC are given as A(1,2,3),B(−1,0,0), and C(0,1,2).
Also, it is given that ∠ABC is the angle between the vectors
BA
and
BC
.
BA
={1−(−1)}
i
^
+(2−0)
j
^
+(3−0)
k
^
=2
i
^
+2
j
^
+3
k
^
BC
={0−(−1)}
i
^
+(1−0)
j
^
+(2−0)
k
^
=
i
^
+
j
^
+2
k
^
∴
BA
⋅
BC
=(2
i
^
+2
j
^
+3
k
^
)⋅(
i
^
+
j
^
+2
k
^
)=2×1+2×1+3×2=2+2+6=10
∣
BA
∣=
2
2
+2
2
+3
2
=
4+4+9
=
17
∣
BC
∣=
1+1+2
2
=
6
Now, it is known that:
BA
⋅
BC
=∣
BA
∣∣
BC
∣cos(∠ABC)
∴10=
17
×
6
cos(∠ABC)
⇒cos(∠ABC)=
17
×
6
10
⇒∠ABC=cos
−1
(
102
10
)
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