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The area of the triangle formed by the points whose position vectors are $3 i + j, 5 i + 2 j + k, i − 2 j + 3 k$ .

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23 sq units

21 sq units

29 sq units

33 sq units

answer is C.

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Detailed Solution

According to the question, we are given the position vectors of three points and we have to calculate the area of the triangle.
Let us assume that A, B, and C are the points.
The position vector of point A =

A → = 3 i + j

………………………….(1)
The position vector of point B =

B → = 5 i + 2 j + k

…………………….(2)
The position vector of point C =

C → = i − 2 j + 3 k

___________________(3)
Here, we are asked to find the area of ΔABC.

https://www.vedantu.com/question-sets/38edd021-5393-4350-9f4f-59c296336cd35297749456963442968.png

We know the formula that if we have three vectors

A →

B →

and

C →

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

A → − C →

and

B → − C →

i.e.,
The area of ΔABC

= 12 B → − C → × A → − C →

………………………...(4)
From equation (1), and equation (3), we get

(A → − C →) = (3 i + j) − (i − 2 j + 3 k) = (3 − 1) i + (1 + 2) j − 3 k = 2 i + 3 j − 3 k

………………….(5)
Similarly, from equation (2), and equation (3), we get

(B → − C →) = (5 i + 2 j + k) − (i − 2 j + 3 k) = (5 − 1) i + (2 + 2) j + (1 − 3) k = 4 i + 4 j − 2 k

……………..(6)
Now, from equation (4), equation (5), and equation (6), we get
The area of

Δ A B C = 12 × [(2 i + 3 j − 3 k) × (4 i + 4 j − 2 k)]

………………..(7)
We know the property that

i × i = 0, j × j = 0, k × k = 0, i × j = k, i × k = − j, j × i = − k

j × k = i, k × i = j, k × j = − i

………………….(8)
Now, using equation (8) and on simplifying equation (7), we get
The area of

Δ A B C = 12 × [6 i − 8 j − 4 k] = 3 i − 4 j − 2 k

…………….(9)
We know the formula for the magnitude of a vector

x i + y j + z k

, Magnitude

= x 2 + y 2 + z 2

Now, from equation (9) and equation (10), we get
The area of

Δ A B C = 32 + (− 4) 2 + (− 2) 2 = 9 + 16 + 4 = 29 s q

units.
Therefore, the area of the triangle is

29 s q

units.

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