Question

find the area of triangle with vertices
(1,2,0), (1;0, 2) and (0,3,1).​

Answer 1

Answer:

The vertices of triangle ABC are given as A(1,1,2),B(2,3,5) and C(1,5,5).

The adjacent sides

AB

and

BC

of ΔABC are given as:

AB

=(2−1)

i

^

+(3−1)

j

^

+(5−2)

k

^

=

i

^

+2

j

^

+3

k

^

BC

=(1−2)

i

^

+(5−3)

j

^

+(5−5)

k

^

=−

i

^

+2

j

^

Area of ΔABC=

2

1

∣

AB

×

BC

∣

AB

×

BC

=

∣

∣

∣

∣

∣

∣

∣

∣

i

^

1

−1

j

^

2

2

k

^

3

0

∣

∣

∣

∣

∣

∣

∣

∣

∴∣

AB

×

BC

∣=

(−6)

2

+(−3)

2

+4

2

=

36+9+16

=

61

Hence, the area of ΔABC is

2

61

square units

Step-by-step explanation:

plzzzzz change the value

Answer 2

The correct answer is $\sqrt{6}$.

Given: Points = (1,2,0), (1,0, 2) and (0,3,1).​

To Find: Area of triangle.

Solution:

The area of triangle = $\frac{1}{2} |AB*AC|$

AB = Coordinates of (B - A)

AB = (0, -2, 2)

AC = Coordinates of (C - A)

AC = (-1, 1, 1)

|AB x AC|

= $\left[\begin{array}{ccc}i&j&k\\0&-2&2\\-1&1&1\end{array}\right]$

= i(-2-2) -j(0-(-2)) + k(0-2)

= -4i -2j - 2k

|-4i -2j - 2k| = $\sqrt{(-4)^{2}+(-2)^{2} + (-2)^{2} }$

= $\sqrt{24} =2\sqrt{6$

$\frac{1}{2} |AB*AC|$ = $\frac{1}{2}*2\sqrt{6}$

= $\sqrt{6}$

Hence, the area of triangle is $\sqrt{6}$.

#SPJ3