Question

The area of the triangle formed by the points whose position vectors are $$3i+j$$ , $$5i+2j+k$$ , $$i-2j+3k$$ .
(A) $$\sqrt{23}$$ sq units
(B) $$\sqrt{21}$$ sq units
(C) $$\sqrt{29}$$ sq units
(D) $$\sqrt{33}$$ sq units

Answer 1

Hint: The position vectors of point A, B, and C are $$3i+j$$ , $$5i+2j+k$$ , and $$i-2j+3k$$ . We know the formula that if we have three vectors $$\overrightarrow{A}$$ , $$\overrightarrow{B}$$ , and $$\overrightarrow{C}$$ . Then, the area of $$\mathrm{\Delta }ABC$$ is given by the half of the vector product of $$(\overrightarrow{A}-\overrightarrow{C})$$ and $$(\overrightarrow{B}-\overrightarrow{C})$$ . Use this formula and calculate the area in vector form. Now, get the magnitude of the area vector using the formula that magnitude of a vector $$xi+yj+zk$$ is $$\sqrt{x^2+y^2+z^2}$$ .

Complete step by step answer:
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 = $$\overrightarrow{A}=3i+j$$ …………………………………………….(1)
The position vector of point B = $$\overrightarrow{B}=5i+2j+k$$ …………………………………………….(2)
The position vector of point C = $$\overrightarrow{C}=i-2j+3k$$ …………………………………………….(3)
Here, we are asked to find the area of $$\mathrm{\Delta }ABC$$ .

We know the formula that if we have three vectors $$\overrightarrow{A}$$ , $$\overrightarrow{B}$$ , and $$\overrightarrow{C}$$ . Then, the area of $$\mathrm{\Delta }ABC$$ is given by the half of the vector product of $$(\overrightarrow{A}-\overrightarrow{C})$$ and $$(\overrightarrow{B}-\overrightarrow{C})$$ i.e.,
The area of $$\mathrm{\Delta }ABC$$ = $${\displaystyle \frac{1}{2}}[(\overrightarrow{A}-\overrightarrow{C})\times (\overrightarrow{B}-\overrightarrow{C})]$$ ………………………………..(4)
From equation (1), and equation (3), we get
$$(\overrightarrow{A}-\overrightarrow{C})=(3i+j)-(i-2j+3k)=(3-1)i+(1+2)j-3k=2i+3j-3k$$ ………………………………………….(5)
Similarly, from equation (2), and equation (3), we get
$$(\overrightarrow{B}-\overrightarrow{C})=(5i+2j+k)-(i-2j+3k)=(5-1)i+(2+2)j+(1-3)k=4i+4j-2k$$ ………………………………………….(6)
Now, from equation (4), equation (5), and equation (6), we get
The area of $$\mathrm{\Delta }ABC$$ = $${\displaystyle \frac{1}{2}}\times [(2i+3j-3k)\times (4i+4j-2k)]$$ ……………………………………….(7)
We know the property that $$i\times i=0$$ , $$j\times j=0$$ , $$k\times k=0$$ , $$i\times j=k$$ , $$i\times k=-j$$ , $$j\times i=-k$$ ,
$$j\times k=i$$ , $$k\times i=j$$ , and $$k\times j=-i$$ ……………………………………..(8)
Now, using equation (8) and on simplifying equation (7), we get
The area of $$\mathrm{\Delta }ABC$$ = $${\displaystyle \frac{1}{2}}\times [6i-8j-4k]$$ = $$3i-4j-2k$$ ……………………………………….(9)
We know the formula for the magnitude of a vector $$xi+yj+zk$$ , Magnitude = $$\sqrt{x^2+y^2+z^2}$$ …………………………………………………(10)
Now, from equation (9) and equation (10), we get
The area of $$\mathrm{\Delta }ABC$$ = $$\sqrt{3^2+{(-4)}^2+{(-2)}^2}=\sqrt{9+16+4}=\sqrt{29}$$ sq units.
Therefore, the area of the triangle is $$\sqrt{29}$$ sq units.
Hence, the correct option is C .

Note:
For solving this type of question, one must remember the formula for the area of the triangle. That is, the area of the triangle enclosed by the position vectors $$\overrightarrow{A}$$ , $$\overrightarrow{B}$$ , and $$\overrightarrow{C}$$ is equal to the vector product of $$(\overrightarrow{A}-\overrightarrow{C})$$ and $$(\overrightarrow{B}-\overrightarrow{C})$$ .