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Question: If \( a = 2i - 3j + k,\;{\text{b = - i + k, c = 2j - k,}} \) find the area of the parallelogram havi...

If a=2i−3j+k,  b = - i + k, c = 2j - k,a = 2i - 3j + k,\;{\text{b = - i + k, c = 2j - k,}} find the area of the parallelogram having diagonals a+ba + b and b+cb + c
A.Area =1421= \dfrac{1}{4}\sqrt {21}
B.Area =1419= \dfrac{1}{4}\sqrt {19}
C.Area =1219= \dfrac{1}{2}\sqrt {19}
D.Area =1221= \dfrac{1}{2}\sqrt {21}

Explanation

Solution

Hint : First of all we will assume two diagonals then will find the diagonals then will place the value of diagonals in the area formula. Will use a cross-product method to find the diagonal product and then will find magnitude.

Complete step-by-step answer :
Let us take the given vertices –
a=2i−3j+k,  b = - i + k, c = 2j - k,a = 2i - 3j + k,\;{\text{b = - i + k, c = 2j - k,}}
Let us assume that diagonals d1=a+b{d_1} = a + b and d2=b+c{d_2} = b + c
Now, given that the diagonals of the parallelogram are –
Find a+ba + b
Place the values in the above expression –
a+b=(2i−3j+k)+(−i+k)a + b = (2i - 3j + k) + ( - i + k)
Open the brackets and simplify. Remember when there is a positive sign outside the bracket then there is no change in the signs of the terms inside the bracket when you open it.
a+b=2i−3j+k−i+ka + b = 2i - 3j + k - i + k
Make the pair of like terms –
⇒a+b=2i−i‾−3j+k+k‾\Rightarrow a + b = \underline {2i - i} - 3j\underline { + k + k}
Simplify the above equation –
⇒a+b=i−3j+2k\Rightarrow a + b = i - 3j + 2k
⇒d2=i−3j+2k\Rightarrow {d_2} = i - 3j + 2k .... (A)
Similarly for second diagonal
b+c=−i+k+2j−kb + c = - i + k + 2j - k
Make the pair of like terms –
b+c=−i+2j−k+k‾b + c = - i + 2j\underline { - k + k}
Terms with the same value and opposite sign cancel each other. Simplify the above equation –
b+c=−i+2jb + c = - i + 2j
Therefore, d2=−i+2j{d_2} = - i + 2j .... (B)
Now, the area of the parallelogram is
A=12∣d1×d2∣A = \dfrac{1}{2}\left| {{d_1} \times {d_2}} \right|
Place values form equation (A) and (B)
A=12∣(i−3j+2k)×(−i+2j)∣A = \dfrac{1}{2}\left| {(i - 3j + 2k) \times ( - i + 2j)} \right| ..... (C)
Now find cross product –
\left| {\begin{array}{*{20}{c}} i&j;&k; \\\ 1&{ - 3}&2 \\\ { - 1}&2&0 \end{array}} \right|
Open the determinant –
=i(−4)−j(2)+k(2−3)= i( - 4) - j(2) + k(2 - 3)
Simplify the above equation – when you subtract a bigger number from smaller there will be a negative sign in resultant value.
=4i−2j−k= 4i - 2j - k
Now find the mode of the above vector expression –
∣4i−2j−k∣=42+(−2)2+(−1)2\left| {4i - 2j - k} \right| = \sqrt {{4^2} + {{( - 2)}^2} + {{( - 1)}^2}}
Remember the square of the negative number also gives the positive value. Since minus multiplied with minus gives plus.
∣4i−2j−k∣=16+4+1\left| {4i - 2j - k} \right| = \sqrt {16 + 4 + 1}
Simplify the above equation –
∣4i−2j−k∣=21\left| {4i - 2j - k} \right| = \sqrt {21} .... (D)
Place above value in equation (C)
⇒A=212\Rightarrow A = \dfrac{{\sqrt {21} }}{2} Sq.units
So, the correct answer is “Option D”.

Note : Parallelogram law:
If two vectors are represented by two adjacent sides of a parallelogram, then the diagonal of parallelogram through the common point represents the sum of the two vectors in both magnitude and direction.