Using vectors, find the area of the triangle with vertices: A(1, 1, 2), B(2, 3, 5), and C(1 , 5, 5) which is known as ____

Question

Infinity Learn · Accepted Answer

The coordinate of point A = (1,1,2) ………………………………….(1)
The coordinate of point B = (2,3,5) ……………………………………..(2)
The coordinate of point C = (1,5,5) ………………………………………..(3)
Now, converting the above points into position vectors by adding

\hat{i}

,

\hat{j}

, and

\hat{k}

in x, y, and z coordinates.
The position vector of A = (1

\hat{i}

+1

\hat{j}

+2

\hat{k}

)…………………………………..(4)
The position vector of B = (2

\hat{i}

+3

\hat{j}

+5

\hat{k}

)…………………………………..(5)
The position vector of C = (1

\hat{i}

+5+5

\hat{k}

)…………………………………..(6)
Question Image

We know the formula that if we have three vectors

A \to, B \to . and C \to

.Then, the area of ΔABC is given by the half of the vector product of (

A \to - C \to

.) and (

B \to - C \to)

. i.e.,
The area of ΔABC =

\frac{1}{2}

[ (

A \to - C \to

)×(

B \to - C \to

)]………………………………..(7)
From equation (4), and equation (6), we get
(

A \to - C \to

) = (1

\hat{i}

+1

\hat{j}

+2

\hat{k}

)−(1

\hat{i}

+5

\hat{j}

+5

\hat{k}

)
= (1−1)i+(1−5)+(2−5)k
= −4

\hat{j}

−3

\hat{k}

…………………………………….(8)
Similarly, from equation (2), and equation (3), we get
(

B \to - C \to)

= (2

\hat{i}

+3

\hat{j}

+5

\hat{k}

)−(1

\hat{i}

+5

\hat{j}

+5

\hat{k}

)
= (2−1)i+(3−5)j+(5−5)k
=1

\hat{i}

−2

\hat{j}

………………………………………….(9)
Now, from equation (7), equation (8), and equation (9), we get
The area of ΔABC =

\frac{1}{2}

×[(−4

\hat{i}

−3

\hat{k}

)×(1

\hat{i}

−2

\hat{j}

)] ………………………………….(10)
We know the property that

\hat{i}

×

\hat{j}

=0 ,

\hat{j}

×

\hat{j}

=0 ,

\hat{k}

×

\hat{k}

=0 ,

\hat{i}

×

\hat{j}

=

\hat{k}

,

\hat{i}

×

\hat{k}

=−

\hat{j}

,

\hat{j}

×

\hat{i}

=-

\hat{k}

,

\hat{j}

×

\hat{k}

=

\hat{i}

,

\hat{k}

×

\hat{i}

=

\hat{j}

, and

\hat{k}

×

\hat{j}

=

\hat{- i}

……………………………………..(11)
Now, using equation (11) and on simplifying equation (10), we get
The area of ΔABC=

\frac{1}{2}

×[−6

\hat{i}

−3

\hat{j}

+4

\hat{k}

]
= −3

\hat{i}

−

\frac{3}{2} \hat{j}

+2

\hat{k}

……………………………………….(12)
We know the formula for the magnitude of a vector xi+yj+zk ,
Magnitude =

\sqrt{x^{2} + y^{2} + z^{2}}

…………………………………………………(13)
Now, from equation (12) and equation (13), we get
The area of ΔABC =

\sqrt{{(- 3)}^{2} + {(- \frac{3}{2})}^{2} + 2^{2}}

=

\sqrt{9 + \frac{9}{4} + 4}

=

\sqrt{\frac{36 + 9 + 16}{4}}

=

\frac{\sqrt 61}{2}

sq units.
Therefore, the area of the triangle is

\frac{\sqrt 61}{2}

sq units.

Using vectors, find the area of the triangle with vertices: A(1, 1, 2), B(2, 3, 5), and C(1 , 5, 5) which is known as ____

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